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Counting |
Now that all of you know what a number is, we will learn to count with them. You already know how to count: 1, 2, 3, ... But do you know the shortcuts?
How many ways can a squirrel climb this tree to the very top? We could count all the green branches: 1, 2, 3, ..., 34, 35, 36. Or we could take a shortcut: 4+4+4+4+4+4+4+4+4 = 36. Can you think of a faster way? If so, you are on your way to finding today's rule.
Click here for the Counting Video.
ASSIGNMENT
Pick one problem from the list below that no one else has picked. THE FIRST 10 PROBLEMS ARE RESERVED FOR LATE COMERS, AND MAY NOT BE POSTED UNTIL ALL THE OTHER PROBLEMS ARE CLAIMED. Some problems are grouped, but pick only one problem number. For 2 points, answer the question in a comment below. Show us how you got your answer.
YOU MAY NOT USE ANY SYMBOLS IN YOUR PROBLEM EXCEPT DIGITS (0-9) AND THE 4 OPERATORS + - * /
DO NOT USE EXPONENT SYMBOLS LIKE ^ . EXPLAIN ALL SOLUTIONS WITH ONLY + - * /
NO FUNNY SYMBOLS FROM FORMULAS YOU COPIED FROM SOMEWHERE ARE ALLOWED.
THIS WAY WE WILL ALL UNDERSTAND YOUR THINKING, AND THERE WILL BE LESS CONFUSION.
If you draw a flower in the applet that illustrates your answer, and add it to your comment, you earn 3 extra points.
To illustrate your answer your drawing must contain these properties:
1. The number of tips on your flower must equal the answer to your problem.
2. Each number in your STEM box must correspond to a number in your calculation, and vice versa.
3. Each number in your calculation must be represented by a different color in your diagram.
Include the input boxes with your drawing so we know what you did to draw your flower.
If you experiment with, and thereby change, the radii and colors, you will earn another point.
WARNING! Some flowers may take a long time to draw, but the view may be worth the wait.
If your computer crashes because it is drawing too many stems,
you may reduce the number of stems by reducing the number of radii;
but show all the numbers in the stem box that you used in your calculation.
If your flower, or tree, pleases me aesthetically, you will receive an extra point.
After you have solved your problem look at your neighbors' solutions.
Can you write a rule for all of these problems that makes counting faster?
Discuss your rule with others.
If you can write your rule correctly in English, you will receive 2 more points.
Obviously, there are not enough problems listed for everyone.
After all problems are taken, I will post an extra problem for really latecomers to solve.
When you start a problem, you claim it for one week only.
You must work on it within a week, or someone else can claim it by solving it.
MR. OLLO'S CLOSET
01. Mr. Ollo stared into his closet and found only two pairs of pants—a light pair and a dark pair—that he could still fit into after the semester break. There he also found five shirts that were not in the laundry basket. How many combinations of pants and shirts can Mr. Ollo wear to school? Verify your answer with a list of all possibilities.
02. You go to your closet and find three tops (blouses or shirts) and four bottoms (skirts or pants) that you can wear. How many outfits can you make? Verify your answer with a list of all possibilities.
03. Mr. Ollo decides to impress his students by wearing a tie. He finds six psychedelic ties that truly amaze the eye. How many combinations of pants, shirts, and ties can he wear?
04. You find five pairs of shoes in your closet that you can wear with your tops and bottoms. How many combinations of tops, bottoms, and shoes can you wear?
05. How many outfits can you wear if you have A number of tops, B number of bottoms, and C number of shoes?
PAINTING A BUILDING
Buildings on Looloololo are mutli-colored; but zoning laws say only one color per floor.
06. How many ways can you paint a 4-story building if each floor can be only one color, and you only have 2 colors of paint?
07. How many ways can you paint a 5-story building if each floor can be only one color, and you only have 2 colors of paint?
08. How many ways can you paint an n-story building if each floor can be only one color, and you only have 2 colors of paint?
09. How many ways can you paint a 4-story building if each floor can be only one color, and you only have 3 colors of paint?
10. How many ways can you paint an n-story building if each floor can be only one color, and you have k colors of paint?
HOW MANY WAYS CAN YOU...
11. ...walk from the center of a wild flower with 8 stems to the end of a flower petal, if each stem has 5 petals? (Pretend you are a bug living in the middle of the flower.)
12. ...click the light switches on and off in your home? (First, count your light switches.)
13. ...roll three polyhedral dice with 4, 6, and 8 sides, respectively?
14. ...answer a True/False test with 20 questions?
15. ...make regular Tennessee license plates?
16. ...make 7 digit phone numbers if the first digit cannot be a 0 or a 1?
17. ...make DNA with 4 bases G A C T and 6 billion slots to place them in?
(Hint: Perhaps you should decrease the 6 billion slots to just 6 when illustrating this one.)
18. ...choose 3 of the 24 Greek letters for a fraternity or sorority name?
19. ...elect a committee of 50 people if each state may appoint either its governor or one of its two senators?
20. ...assign a variable a name with 8 characters in a C program if a variable name may contain uppercase letters, lowercase letters, digits, and underscores? However, the first character may not be a digit?
21. ...assign cell phone numbers if each number has 10 digits in this form: NXX-NXX-XXXX? The first 3 digits are the country's prefix. The X represents any digit from 0 thru 9. The N represents any digit from 2 thru 9. Could everyone alive now have their own phone number?
22. ...make 8 letter Looloolo words? Looloolo is one of them?
23. ...watch a team win or lose the first four games of a World Series?
24. ...order a drink from an ice cream parlor that offers three drinks, sodas, milk shakes, and blasters in three sizes, small, medium, and large, with 24 flavors. If you order one drink, how many choices do you have?
25. ...take DNA strands of 6 acids in length from an alien race whose DNA consists of 5 amino acids?
26. ...take DNA strands of 5 acids in length from an alien race whose DNA consists of 6 amino acids?
THE SLOT MACHINE
The Twenty-One Bell slot machine has three reels with 20 stop positions on each reel. Most of the positions have one symbol but a few positions have two. The arrangement of the symbols is shown in the table below:
27. The smallest payoff is for a cherry on reel 1 with no cherries on reels 2 and 3.
How many ways can you win this payoff?
28. The next smallest payoff is for cherries on the first two reels. How many ways can you win this payoff?
29. One of the jackpot payoffs is for melons on reels 1 and 2 and a bar on reel 3.
How many ways can you win this jackpot?
30. Another way to win the jackpot is with three melons. How many ways can you win this jackpot?
31. The next way to win the jackpot is with three bars. How many ways can you win this jackpot?
32. The last way to win the jackpot is with three 7's. How many ways can you win this jackpot?
33. Another payoff is oranges on reels 1 and 2 and a bar on reel 3. How many ways can you win this payoff?
34. Another payoff is three oranges. How many ways can you win this payoff?
35. Another payoff is plums on reels 1 and 2 and a bar on reel 3. How many ways can you win this payoff?
36. Another payoff is three plums. How many ways can you win this payoff?
37. Another payoff is bells on reels 1 and 2 and a bar on reel 3. How many ways can you win this payoff?
38. Another payoff is three bells. How many ways can you win this payoff?
39. In how many different positions can the three reels of the slot machine stop?
Comments:
From wHolt - 12/12/06 12:09 PM
From BassLady - 12/11/06 8:21 AM
#38 - Another payoff is three bells. How many ways can you win this payoff? 40 ways.
From wHolt - 12/9/06 10:39 AM
Kathi - in your stems box i see a 2 and an 8.
but not all those 2's.
make your drawing so it shows all your 2's.
Thanks.
From Kathi - 12/8/06 8:38 AM
2^n where n = number of games.
Sincer there are only 4 games n = 4.
2 *2 * 2 * 2 = 16
From wHolt - 12/7/06 11:16 AM
DirtyBird - what does the second column refer to in your table?
| SWITCH A | what does this column refer to? |
| ON | OFF |
| OFF | ON |
List all the possibilities for your 3 switches in one table.
Kathi - if you were to write your solution without the ^,
how would you do it?
From DirtyBird - 12/6/06 7:18 PM
| SWITCH A | |
| ON | OFF |
| OFF | ON |
| SWITCH B | |
| OFF | ON |
| ON | OFF |
| SWITCH C | |
| ON | OFF |
| OFF | ON |
There is not a medium state with power. You are either on or off. I dont see where you are saying there is 3 sources to the first switch. Thanks.
From Kathi - 12/6/06 6:44 AM
From wHolt - 12/5/06 11:40 AM
Kathi - I cannot argue with a list that lists all the possibilities,
and since your diagram is based on your lists, I will accept it also.
But if you were not making a list,
what shortcut would you use to calculate your solution?
From Kathi - 12/4/06 9:38 AM
I can only find 16 ways that a team can win or loose the first four games.
If the first team looses the first game there are 8 combinations in which the first game is lost. If the first team wins there is 8 combinations in which the first game is won.
These tables show all possibilities if they lose the first game or if they win the first game.
| 1 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 |
| 1 | 1 | 1 | 1 |
| 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 0 | 1 | 1 | 1 |
From BassLady - 12/4/06 9:29 AM
I have not finished my diagram yet. I am working on the radii. I have the stems figured out but not the other part.
From wHolt - 12/3/06 1:11 PM
Was there a faster way?
Did you draw us a picture yet?
Your flower or tree should show 1 choice for the first reel,
5 choices for the 2nd, and 8 for the 3rd?
From BassLady - 12/2/06 8:18 PM
#38 - Another payoff is 3 bells. How many ways can you win this payoff?
There are 40 different ways to win.
| Reel 1 | Reel 2 | Reel 3 | Reel 1 | Reel 2 | Reel 3 | Reel 1 | Reel 2 | Reel 3 | ||
| 8 | 6 | 1 | 8 | 8 | 1 | 8 | 19 | 1 | ||
| 8 | 6 | 4 | 8 | 8 | 4 | 8 | 19 | 4 | ||
| 8 | 6 | 7 | 8 | 8 | 7 | 8 | 19 | 7 | ||
| 8 | 6 | 9 | 8 | 8 | 9 | 8 | 19 | 9 | ||
| 8 | 6 | 12 | 8 | 8 | 12 | 8 | 19 | 12 | ||
| 8 | 6 | 14 | 8 | 8 | 14 | 8 | 19 | 14 | ||
| 8 | 6 | 17 | 8 | 8 | 17 | 8 | 19 | 17 | ||
| 8 | 6 | 19 | 8 | 8 | 19 | 8 | 19 | 19 | ||
| 8 | 11 | 1 | 8 | 14 | 1 | |||||
| 8 | 11 | 4 | 8 | 14 | 4 | |||||
| 8 | 11 | 7 | 8 | 14 | 7 | |||||
| 8 | 11 | 9 | 8 | 14 | 9 | |||||
| 8 | 11 | 12 | 8 | 14 | 12 | |||||
| 8 | 11 | 14 | 8 | 14 | 14 | |||||
| 8 | 11 | 17 | 8 | 14 | 17 | |||||
| 8 | 11 | 19 | 8 | 14 | 19 | |||||
From wHolt - 12/2/06 12:49 PM
but the first 10 problems are off limits till all the other problems are claimed
From Cheana - 12/1/06 8:40 PM
#4, You find five pairs of shoes in your closet that you can wear with your tops and bottoms. How many combinations of tops, bottoms, and shoes can you wear?
5*3*4= 60
From wHolt - 11/30/06 9:36 AM
Soller - thanks for your contribution,
but problems 1 thru 10 are off limits till all the others are solved.
Pick one that is not claimed.
Melewen - sorry about the mac incompatibility.
However, the colleges are lazy
because they have not updated their Java plug-in for a few years.
From Soller - 11/29/06 7:36 PM
...3 shirts and 4 bottoms, how many outfits could I make?
4*3=12
From Melewen - 11/28/06 5:56 PM
From wHolt - 11/27/06 9:23 AM
From Melewen - 11/26/06 10:44 AM
My computer chops off the little text boxes, but stems says 1, 1, 1
From Melewen - 11/26/06 10:07 AM
32. The last way to win the jackpot is with three 7's. How many ways can you win this jackpot?
There is only one 7 in the first column, so the options of getting a 7 in that slot is naturally only one. Similarly there's only one 7 in the second column, so again that possibility is one. It's the same for the third column. Thus, to find how many ways you can win this jackpot, you multiply 1 x 1 x 1 = 1
There is only one way to win this jackpot.
From wHolt - 11/23/06 10:58 AM
From BassLady - 11/22/06 11:48 PM
From wHolt - 11/20/06 9:53 AM
DirtyBird - I noticed that your diagram had 3 green stems.
This means that at some point you would have 3 choices for a some switch.
If that is not so, then revise your diagram.
I edited the rules for diagrams in the instructions.
Perhaps it will clarify a bit.
If not tell me what I should change. Thanks.
From DirtyBird - 11/19/06 8:09 PM
From wHolt - 11/17/06 2:00 PM
Read how the others are counting in their problem.
Assume Mr. Ollo decided his wardrobe each day by spinning the reels on a slot machine.
He had his shoes listed on reel 1, pants on reel 2, and shirts on reel 3.
The question might be: how many outfits can he wear with his dark pair of pants?
From BassLady - 11/16/06 6:20 PM
# 38 = If one needs 3 bells to win then there is only one way to win off of one spin
From BassLady - 11/16/06 6:11 PM
38. Another payoff is three bells. How many ways can you win this payoff?
There is only one way he can win. If he needs 3 bells and there is only 3 positions - he only has one way to win.
From wHolt - 11/16/06 10:36 AM
From Fro - 11/15/06 6:45 PM
31. The next way to win the jackpot is with three bars. How many ways can you win this jackpot?
There are 3 bars in the first row (green), 2 bars in the second (blue) and 1 in the third row (red).
That makes a total of 6 ways to win.
3*2*1=6
From wHolt - 11/15/06 11:58 AM
From BassLady - 11/14/06 7:53 PM
MR. OLLO'S CLOSET
01. Mr. Ollo stared into his closet and found only two pairs of pants—a light pair and a dark pair—that he could still fit into after the semester break. There he also found five shirts that were not in the laundry basket. How many combinations of pants and shirts can Mr. Ollo wear to school? Verify your answer with a list of all possibilities.
He can wear 10 different outfits. Lets say his shirts are pink, yellow, purple, red, and orange. His 2 pair of pants are blue and tan.
#1 - Tan pants and pink shirt, #2 - Tan pants with yellow shirt, #3 - Tan pants and purple shirt, #4 - Tan pants and red shirt, #5 is tan pants with orange shirt.
#6 - blue pants with pink shirt, #7 - blue pants wth yellow shirt, #8 - blue pants with purple shirt, #9 - blue pants and red shirt, and #10 is blue pants with orange shirt.
From wHolt - 11/14/06 12:03 AM
Does the light switch have a state between on and off, perhaps a medium state?
Since you only have 3 switches, list all 6 possibilities.
That's the sure way to convince us.
From DirtyBird - 11/13/06 6:55 PM
12. ...click the light switches on and off in your home?
3 switches that move 2 ways= 6 ways
From wHolt - 11/13/06 11:37 AM
GolfGirl - nice list, but how does it relate to your flower diagram?
Also how would you compute the number of outfits without making a list?
Also, the first 10 problems are not available till after all the others are taken.
Trixie - thanks for the list and diagram, however
the first 10 problems are not available till after all the others are taken.
So pick another one not taken yet.
Although, many of you are using the same method to figure your solutions,
no one has ventured, as yet, to state the general rule for your short cut counting method.
Remember, you are trying to find a general rule
that will apply to all the remaining problems in this course.
From Zonino - 11/12/06 10:37 PM
28. The next smallest payoff is for cherries on the first two reels. How many ways can you win this payoff?
In order to win this payoff, the first two reels must be cherries and there must be no cherries on the third reel. Since there are no cherries on the third reel to begin with, this means that all 20 stops are possible. For Reel 1, there are two possible stops that have cherries and for reel 2 there are 6 possible stops that have cherries.
Therefore, there would be 2*6*20 ways to have cherries on the first two reels or 240 stems.
From SkoolGirl - 11/12/06 7:52 PM
34. Another payoff is three oranges. How many ways can you win this payoff?
The green stems represent the 5 slots with oranges on the 1st reel.
The blue stems represent the 5 slots with oranges on the 2nd reel.
The red stems represent the 4 slots with oranges on the 3rd reel.
5*5*4=100
Their are 100 ways to win the jackpot with 3 oranges.
From Draco - 11/12/06 7:10 PM
30. Another way to win the jackpot is with three melons. How many ways can you win this jackpot?
There are eight ways to win the jackpot with three melons.
The green stems represent the first slot which contains two melons, the blue stems represent the second slot which also contains two melons, and the red stems represent the third slot which cotains two melons. (2*2*2=8)
From Trixie - 11/12/06 3:53 PM
# 2
If you have 3 different tops and 4 different bottoms you have a total of 12 possible outfits to wear. (3 * 4 = 12)
| Top 1 | Bottom 1 |
| Top 1 | Bottom 2 |
| Top 1 | Bottom 3 |
| Top 1 | Bottom 4 |
| Top 2 | Bottom 1 |
| Top 2 | Bottom 2 |
| Top 2 | Bottom 3 |
| Top 2 | Bottom 4 |
| Top 3 | Bottom 1 |
| Top 3 | Bottom 2 |
| Top 3 | Bottom 3 |
| Top 3 | Bottom 4 |
From GolfGirl - 11/12/06 2:41 PM
problem #1 | shirt # | Pants color |
| 1 | L |
| 2 | L |
| 3 | L |
| 4 | L |
| 5 | L |
| 1 | D |
| 2 | D |
| 3 | D |
| 4 | D |
| 5 | D |
From wHolt - 11/12/06 2:22 PM
BassLady - Pringle already completed your choice. Pick another.
Sushine - are there only 2 choices for the second digit?
Why are there only 2 blue stems?
Houdini - you were not required to make such an exhaustive table
but you definitely deserve extra points for it.
Melewen - Sushine beat you to it. Pick another.
Centerfield - you do not need to write the full number out all the way.
From CenterField - 11/12/06 1:44 PM
20. ...assign a variable a name with 8 characters in a C program if a variable name may contain uppercase letters, lowercase letters, digits, and underscores? However, the first character may not be a digit?
Uppercase letters = 26 options
Lowercase letters = 26 options
Digits = 10 options
Underscore = 1 option
Add the options together and you get 63 options for each position, except for the first position (no digits) which has 53 options.
You then multiply the total options for the eight places together…
53 * 63 * 63 * 63 * 63 * 63 * 63 * 63 = 2.08766E14
From Melewen - 11/12/06 10:57 AM
21. ...assign cell phone numbers if each number has 10 digits in this form: NXX-NXX-XXXX? The first 3 digits are the country's prefix. The X represents any digit from 0 thru 9. The N represents any digit from 2 thru 9. Could everyone alive now have their own phone number?
8x10x10x8x10x10x10x10x10x10 = 6,400,000,000
yes! everyone can have their own phone number. i will do the picture when i can get to a computer that will let me!
From Houdini - 11/12/06 12:25 AM
#39 In how many different positions can the three reels of the slot machie stop?
One outcome, from three reels, with 20 different positions for each reel.
20^3 or 20x20x20=8000
There are eight thousand different outcomes.
A table for all outcomes with reel one at stop one, using numbers and letters:
| 111 | 121 | 131 | 141 | 151 | 161 | 171 | 181 | 191 | 101 | 1q1 | 1w1 | 1e1 | 2r1 | 1t1 | 1y1 | 1u1 | 1i1 | 1o1 | 1p1 |
| 112 | 122 | 132 | 142 | 152 | 162 | 172 | 182 | 192 | 102 | 1q2 | 1w2 | 1e2 | 1r2 | 1t2 | 1y2 | 1u2 | 1i2 | 1o2 | 1p2 |
| 113 | 123 | 133 | 143 | 153 | 163 | 173 | 183 | 193 | 103 | 1q3 | 1w3 | 1e3 | 1r3 | 1t3 | 1y3 | 1u3 | 1i3 | 1o3 | 1p3 |
| 114 | 124 | 134 | 144 | 154 | 164 | 174 | 184 | 194 | 104 | 1q4 | 1w4 | 1e4 | 1r4 | 1t4 | 1y4 | 1u4 | 1i4 | 1o4 | 1p4 |
| 115 | 125 | 135 | 145 | 155 | 165 | 175 | 185 | 195 | 105 | 1q5 | 1w5 | 1e5 | 1r5 | 1t5 | 1y5 | 1u5 | 1i5 | 1o5 | 1p5 |
| 116 | 126 | 136 | 146 | 156 | 166 | 176 | 186 | 196 | 106 | 1q6 | 1w6 | 1e6 | 1r6 | 1t6 | 1y6 | 1u6 | 1i6 | 1o6 | 1p6 |
| 117 | 127 | 137 | 147 | 157 | 167 | 177 | 187 | 197 | 107 | 1q7 | 1w7 | 1e7 | 1r7 | 1t7 | 1y7 | 1u7 | 1i7 | 1o7 | 1p7 |
| 118 | 128 | 138 | 148 | 158 | 168 | 178 | 188 | 198 | 108 | 1q8 | 1w8 | 1e8 | 1r8 | 1t8 | 1y8 | 1u8 | 1i8 | 1o8 | 1p8 |
| 119 | 129 | 139 | 149 | 159 | 169 | 179 | 189 | 199 | 109 | 1q9 | 1w9 | 1e9 | 1r9 | 1t9 | 1y9 | 1u9 | 1i9 | 1o9 | 1p9 |
| 110 | 120 | 130 | 140 | 150 | 160 | 170 | 180 | 190 | 100 | 1q0 | 1w0 | 1e0 | 1r0 | 1t0 | 1y0 | 1u0 | 1i0 | 1o0 | 1p0 |
| 11q | 12q | 13q | 14q | 15q | 16q | 17q | 18q | 19q | 10q | 1qq | 1wq | 1eq | 1rq | 1tq | 1yq | 1uq | 1iq | 1oq | 1pq |
| 11w | 12w | 13w | 14w | 15w | 16w | 17w | 18w | 19w | 10w | 1qw | 1ww | 1ew | 1rw | 1tw | 1yw | 1uw | 1iw | 1ow | 1pw |
| 11e | 12e | 13e | 14e | 15e | 16e | 17e | 18e | 19e | 10e | 1qe | 1we | 1ee | 1re | 1te | 1ye | 1ue | 1ie | 1oe | 1pe |
| 11r | 12r | 13r | 14r | 15r | 16r | 17r | 18r | 19r | 10r | 1qr | 1wr | 1er | 1rr | 1tr | 1yr | 1ur | 1ir | 1or | 1pr |
| 11t | 12t | 13t | 14t | 15t | 16t | 17t | 18t | 19t | 10t | 1qt | 1wt | 1et | 1rt | 1tt | 1yt | 1ut | 1it | 1ot | 1pt |
| 11y | 12y | 13y | 14y | 15y | 16y | 17y | 18y | 19y | 10y | 1qy | 1wy | 1ey | 1ry | 1ty | 1yy | 1uy | 1iy | 1oy | 1py |
| 11u | 12u | 13u | 14u | 15u | 16u | 17u | 18u | 19u | 10u | 1qu | 1wu | 1eu | 1ru | 1tu | 1yu | 1uu | 1iu | 1ou | 1pu |
| 11i | 12i | 13i | 14i | 15i | 16i | 17i | 18i | 19i | 10i | 1qi | 1wi | 1ei | 1ri | 1ti | 1yi | 1ui | 1ii | 1oi | 1pi |
| 11o | 12o | 13o | 14o | 15o | 16o | 17o | 18o | 19o | 10o | 1qo | 1wo | 1eo | 1ro | 1to | 1yo | 1uo | 1io | 1oo | 1po |
| 11p | 12p | 13p | 14p | 15p | 16p | 17p | 18p | 19p | 10p | 1qp | 1wp | 1ep | 1rp | 1tp | 1yp | 1up | 1ip | 1op | 1pp |
To show all possibilities you would need to repeat this table for every stop of reels one, two, and three.
And now, my lovely Christmas-like drawing:
From Sunshine - 11/11/06 8:38 PM
21. ...assign cell phone numbers if each number has 10 digits in this form: NXX-NXX-XXXX? The first 3 digits are the country's prefix. The X represents any digit from 0 thru 9. The N represents any digit from 2 thru 9. Could everyone alive now have their own phone number?
This picture is actually a shortened representation because the number is so large.
I figure n=8 possibilites times 2 places = 8^2
x=10 possibilites times 8 places = 10^8
10^8 x 8^2 = 6400000000
From BassLady - 11/11/06 8:08 PM
3 different drinks, with 3 different sizes and 24 flavors.
Each drink has 72 different varieties. 1 (drink) x (3 sizes) x 24 flavors.
72 x 3 = 216 different drinks can be made.
From wHolt - 11/11/06 7:45 PM
Pac - are you confusing bases with slots?
In the problem 17, there are 4 bases and 6 slots ( for the reduced problem).
How many ways to fill the 1st slot?
How many ways to fill the 2nd slot?
.
.
.
How many ways to fill the 6th slot?
David - nice cherry trees
7Iron - nice binary tree
From 7Iron - 11/11/06 7:08 PM
Here's the tree #22
There are 256 8 letter words in the Looloolo dictionary. There are only two letters in the dictionary. If there are two letters in the dictionary and we are only counting the number if 8 letter words you get 256. Also, Looloolo is included in this, since it is an 8 letter word.
2*2*2*2*2*2*2*2 = 256
From David - 11/11/06 4:45 PM
27. The smallest payoff is for a cherry on reel 1 with no cherries on reels 2 and 3.
How many ways can you win this payoff?
The path towards the smallest payoff (yippee) starts on the third reel, which must be a cherry. There are two cherries on the first reel. If you get one of the two cherries on the first reel, the second reel must not have any cherries. There are 14 stops on that reel without any cherries. finally on the third reel, you must not have any cherries. Good thing there arent any charries on the third reel, so there are 20 positions on the reel without cherries.
Writing it out would be 2*14*20 ways to win the lowest payoff, or 560 paths.
From Pac - 11/11/06 3:09 PM
So thinking about it again, perhaps my original thought was correct even though calculators and Excel won't help me find the answer...
If you have six bases, you have 4 options for each, so you'd have 4*4*4*4*4*4 = 4,096 options for your six bases. If you had 6,000,000,000 bases, your number of possibilities would be so big that I don't think you'd want me to type it here even if I did have the life expectancy to figure it out.
Therefore, here's an updated flower deal with 4,096 options...
From wHolt - 11/10/06 7:25 PM
SuperDuke - you have rediscovered the Sierpinski triangle!
Tiger - you need as many colors as you have stems.
Perhaps try shorter radii.
7Iron - you would draw a tree if you put a 1 in front of all those 2's.
From 7Iron - 11/10/06 9:45 AM
#22
There are 256 8 letter words in the Looloolo dictionary. There are only two letters in the dictionary. If there are two letters in the dictionary and we are only counting the number if 8 letter words you get 256. Also, Looloolo is included in this, since it is an 8 letter word.
2*2*2*2*2*2*2*2 = 256
From Tiger - 11/9/06 9:31 PM
#15 There are 6 places on a license plate.
There are 26 letters in the alphabet and 10 digits for a total of 36 possibilties in each place.
36*36*36*36*36*36=2176782336
From wHolt - 11/8/06 10:58 AM
Boki -
There are only 8 stems.
Where did the other 2 come from?
The bug does not double back.
So dont count the return trip.
How many total paths can the bug walk one way?
Capricorn -
how many acids may you choose from for the first position?
how many for the 2nd?
how many for the 3rd?
how many for the 4th?
how many for the 5th?
or name your acids ABCDEF
and list all 30 possibilities
Taurus -
how many acids may you choose from for the first position?
how many for the 2nd?
how many for the 3rd?
how many for the 4th?
how many for the 5th?
how many for the 6th?
or name your acids ABCDE
and list all 30 possibilities
From Taurus - 11/7/06 6:17 PM
25. ...take DNA strands of 6 acids in length from an alien race whose DNA consists of 5 amino acids?
I will be on an alien DNA strand, which will be made of 6 acids, which could be a combo of 5 amino acids. So there could be 30 different combinations.
From Capricorn - 11/7/06 6:07 PM
26. ...take DNA strands of 5 acids in length from an alien race whose DNA consists of 6 amino acids?
If there is one DNA strand that has 5 acids, but could be of 6 differ acids, then the total amt of combinations per one DNA strand would be 30.
From Boki - 11/7/06 1:11 PM
Thanks for help. Here are paths the bug can take:
if he picks:
1 stem
1111122222333334444455555666667777788888
1234512345123451234512345123451234512345
It will make 40 paths the bug can take which is 8*5
2 stems
1111122222333334444455555666667777788888
1234512345123451234512345123451234512345
1111122222333334444455555666667777788888
1234512345123451234512345123451234512345
It will make 80 paths the bug can take which is 2*8*5
:
:
:
and so on up to
:
8 stems
It will make (1+2+3+4+5+6+7+8)(8*5) paths the bug can take.
(1+2+3+4+5+6+7+8)(8*5)=1440.
In general:
n stems, each with 5 petals
(1+2+.....+n)(n*5)
or
n stems, each with m petals
(1+2+.....+n)(n*m)
From wHolt - 11/7/06 11:54 AM
Boki - that's a lot of bug walking and a lot of brain talking.
but let's make a list of all possible paths.
number all 8 stems and the 5 tips off each stem.
the bug starts in the middle of the flower and has 8 choices.
he picks one and walks to the end where he then has 5 choices:
Here is a list of all the choices:
1 1
1 2
1 3
1 4
1 5
2 1
2 2
2 3
2 4
2 5
3 1
3 2
3 3
3 4
3 5
4 1
4 2
4 3
4 4
4 5
5 1
5 2
5 3
5 4
5 5
6 1
6 2
6 3
6 4
6 5
7 1
7 2
7 3
7 4
7 5
8 1
8 2
8 3
8 4
8 5
Now, where are the other 217 423 888 460 499 999 999 999 999 999 960 paths the bug can walk?
From wHolt - 11/7/06 12:09 AM
Pac -
There are now more than 6 billion people.
When the population reaches 24 billion, will that mean someone will have a clone.
Or when we reach 48 billion, will everyone have a clone?
BASES is a misleading word; but that's what the DNA acid is composed of.
Here is my secret to counting when I am confused. I ask:
How many ways can I fill the first slot?
And for each of those ways, how many ways can I fill the next?
And how many ways can I fill the 3rd slot?
4th?
5th?
...
Poovey - your RADII and COLORS contain only 6 numbers.
This means your flower will only draw 6 levels.
But dont bother to change it. We could not see it anyway.
DirtyBird - assume you had 4 switches in your house.
Could you list all the ways to turn them on?
According to your method, it would be 2x4 = 8
Prove your method works by listing all 8.
If 4 are too many, try 3 switches. Then 2x3=6 ways.
From Boki - 11/6/06 9:17 PM
I guess I needed to multiply all of them.
(8*7*5*4*3*2*1)(5*4*3*2*1) (7*5*4*3*2*1) (5*4*3*2*1) (6*5*4*3*2*1)(5*4*3*2*1) (5*4*3*2*1)(5*4*3*2*1) (4*3*2*1)(5*4*3*2*1) (3*2*1)(5*4*3*2*1) (2*1)(5*4*3*2*1) (1)(5*4*3*2*1)= 2.174238884605 x 10^32
Since10^32=100 000 000 000 000 000 000 000 000 000 000
we will have total arrangements of:
217 423 888 460 500 000 000 000 000 000 000
From DirtyBird - 11/6/06 8:29 PM
12. ...click the light switches on and off in your home?
I have 14 switches in my house that move 2 ways each. So if I used 14*2 then that would give me 28 ways (14*2=28)to turn the switches on or off.
From Poovey - 11/6/06 6:37 PM
16. ...make 7 digit phone numbers if the first digit cannot be a 0 or a 1?
I used 8x10x10x10x10x10x10 because the 8 out of ten digits could be used in all 7 spots. Which totals 8,000,000
From Pac - 11/6/06 2:32 PM
#17: How many ways can you make DNA with 4 bases G A C T and 6 billion slots to place them in?
I tend to overthink things like this, but I think I've figured it out...
At first I thought that 4 bases in 6,000,000,000 slots would be 4x4x4x4x4x4x4x4x4x4x4x (and on and on until you've got 6,000,000,000 4s) which would be a number so large that even Excel refused to display the answer in a whole number format.
After thinking about it, the answer isn't exponential like that. There are four options per base and there are 6,000,000,000 bases. That's 4 x 6,000,000,000 = 24,000,000,000 possible combinations.
I actually tried to get the applet to display that, but I didn't have any luck. I reduced the 6,000,000,000 to 6 per the instructions...
4 options in 6 bases = 4x6 = 24 possible combinations...
From wHolt - 11/6/06 2:00 PM
I will subtract points if you do not explain why you did what you did.
You do not need as much detail as Boki's, but it will impress us if you do.
From wHolt - 11/6/06 1:58 PM
from one tip to all the others without repeating any connection?
But is that the same number of ways a bug can walk from the center to a flower tip?
From wHolt - 11/6/06 1:55 PM
#12 is still open for grabs...
From Trixie - 11/6/06 1:22 PM
From Boki - 11/5/06 9:27 PM
The numbers 40, 39, 38, 37, ... were mistake. I thought only about petals (40 all together) and I didn’t count
on stems at all; therefore, I get previous result which is actually the number of permutations of 40 petals only.
If we have 8 stems and 5 petals, each stem individually has different number of choices.
If we pick first green stem and take an imaginative walk to the end of the stem, we will have (8*7*5*4*3*2*1)(5*4*3*2*1) choices, for2nd green stem we will have (7*5*4*3*2*1) (5*4*3*2*1) choices, for 3rd green stem we
will have (6*5*4*3*2*1)(5*4*3*2*1) choices, for 4th green stem we will have (5*4*3*2*1)(5*4*3*2*1) choices,
for 5th green stem we will have (4*3*2*1)(5*4*3*2*1) choices, for 6th green stem we will have (3*2*1)(5*4*3*2*1) choices, for 7th green stem we will have (2*1)(5*4*3*2*1) choices, and for 8th green stem we will have (1)(5*4*3*2*1).
Let n1, n2, n3,……, nk represent the total number of choices for our individual stems and its petals. By multiplying choices of stems and petals, we will have a total number of choices for each individual
combination as following:
1st green stem with its petals has (8*7*5*4*3*2*1)(5*4*3*2*1) choices: n1=4838400
2nd green stem with its petals has (7*5*4*3*2*1) (5*4*3*2*1) choices: n2=604800
3rd green stem with its petals has (6*5*4*3*2*1)(5*4*3*2*1) choices: n3=86400
4th green stem with its petals has (5*4*3*2*1)(5*4*3*2*1) choices: n4=14400
5th green stem with its petals has (4*3*2*1)(5*4*3*2*1) choices: n5=2880
6th green stem with its petals has (3*2*1)(5*4*3*2*1) choices: n6=720
7th green stem with its petals has (2*1)(5*4*3*2*1) choices: n7=240
8th green stem with its petals has (1)(5*4*3*2*1) choices: n8=120
Now we can add all of them together and get total number of choices by using following formula:
n1 + n2 + … + nr = n
Let n be a total number of choices, then we have:
n=n1 + n2 + n3 + n4 + n5 + n6 + n7 + n8 =4838400 +604800 + 86400 + 14400 + 2880 + 720 + 240 + 120 = 5 547 960
From Harkar - 11/5/06 8:39 PM
#13. ...roll three polyhedral dice with 4, 6, and 8 sides, respectively?
8x6x4= 192
Dice 1 has 8 possibilities; Dice 2 had 6; and Dice 3 has 4, so 8 x 6 x 4 = 192.
From wHolt - 11/5/06 2:54 PM
From JooJoo - 11/5/06 1:54 PM
From SuperDuke - 11/5/06 9:58 AM
#19 is How many ways can you elect a committee of 50 people if each state may appoint either its governor or one of its two senators?
3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3=717897987691852588770249
you will make a choice of one out of the three people for each of the fifty states. This means you have 3 choices on the first state, 9 choices to reach the second state, 27 choices to get to the third state, 81 choices to get to the fourth,. . . when you choose the man for the 50th state you will have exhausted 717897987691852588770249 possible combinations of choices.
My Rendering of this is done with only six states instead of 50 so that we can see the actual choices.
From wHolt - 11/4/06 2:02 PM
From JooJoo - 11/4/06 1:54 PM
18. How many ways can you choose 3 of the 24 Greek letters for a fraternity or soroity name?
24 x 24 x 24= 13,824 ways
Because if you have 3 spots ex:) ABC then A could be 24 different letters and B could be 24 different letters and C could be 24 different letters, because you have 24 letters to choose from in the Greek alphabet.
From wHolt - 11/3/06 11:54 AM
Boki - nice drawing.
You say there are 3,100,796,899,200 paths for the bug to walk
from the center of the flower to one of the 40 tips.
Pick a green stem and take an imaginative walk to the end of the stem.
How many choices do you then have for a blue tip?
Do you have the same choices for any green stem you select?
I do not see the numbers 40, 39, 38, 37, ... mentioned anywhere in the problem.
So I do not know why you are using them in your calculation.
Explain where they came from.
Also , do not use any funny symbols like ! to explain your answer.
All these problems can be answered simply with
addition, subtraction, multiplication, and division.
From wHolt - 11/3/06 11:48 AM
Kathi - thanks for the table.
So you are saying that if you made a table for winning the first game
it would look like this with 8 more possibilities:
| 1 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 |
| 1 | 1 | 1 | 1 |
Now we have yours (8 ways) plus mine (8 more ways) = 16 ways.
Where are the other 16 ways that make 4*8=32 total ways?
From Kathi - 11/3/06 9:07 AM
In order to find all the combinations I had to start with a truth table. Since it ended up being a little big I will show only the options for if the team lost the first game. This gives us the outcomes for one object. Since there are 8 outcomes and 4 objects (games) the total would be 32 different outcomes for a total of 4 games. (8*4=32)
| 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 0 | 1 | 1 | 1 |
From Poovey - 11/2/06 3:55 PM
From wHolt - 11/2/06 11:11 AM
From Boki - 11/2/06 12:19 AM
11. ...walk from the center of a wild flower with 8 stems to the end of a flower petal, if each stem has 5 petals? (Pretend you are a bug living in the middle of the flower.)
Answer:
With 8 stems and 5 petals on each of them I (as a bug) have 40*39*38*37*36*35*34*33
= 3 100 796 899 200 possibilities to walk from the center of a wild flower to the end of a flower petal.
Proof:
If we have 8 stems and 5 petals on each of them, we have 40 pedals in all. In order to find out how many ways is possible to walk from the center of a wild flower to the end of a flower petal, we need to use permutations or multiplicative rule:
or
n1 x n2 x n3 x,……,x nk
In this case n=40 (# of pedals, each of 8 stems has 5 pedals, all together 8*5=40), and r=8 (# of stems)
40P8= 40!/(40-8)!=40!/32!=40*39*38*37*36*35*34*33
= 3 100 796 899 200
Or
For n1 there are 40 choices, for n2 there are 39 choices, for n3 there are 38 choices, for n4 there are 37 choices, for n5 there are 36 choices, for n6 there are 35 choices, for n7 there are 34 choices, and for n8 there are 33 choices.
So,
n1 x n2 x n3 n4 x n5 x n6 x n7 x n8=40*39*38*37*36*35*34*33
= 3 100 796 899 200
From Poovey - 11/1/06 8:56 PM
16. ...make 7 digit phone numbers if the first digit cannot be a 0 or a 1?
There are 10 digits to use, but not with 0 or 1 as the first digits so that is down to 8 to use in 7 different positions
is it 56 different ways.......
From wHolt - 11/1/06 11:42 AM
If Slick's method works for a 20 question test , then it should work for a 3 question test.
So 2*3=6 should be the number of ways to answer a true/false test with 3 questions.
If however we could list a 7th way, we would prove Slick's method incorrect.
If we can only find 6 ways, then we would validate Slick's method,
even though we would not be proving it.
From wHolt - 11/1/06 11:37 AM
Is this a coincidence? Why are they the same number?
From Kathi - 11/1/06 7:47 AM
There are two ways to look at this question. If you are looking at each of the 4 games individually then:
23. ...watch a team win or lose the first four games of a World Series?
4*2 = 8
There are 4 games and either the team wins or loses (2 choices). 4 games * 2 choices = 8 outcomes.
If your team is the Titans:
1st game -- Titans win the game (1st out come) they lose the game (2nd outcome)
2nd game-- Titans win the game (3rd out come) they lose the game (4th outcome)
3rd game--Titans win the game (5th out come) they lose the game (6th outcome)
4th game-- Titans win the game (7th out come) they lose the game (8th outcome)
If you are looking at the total of the 4 games and if the team will win then there are 5 outcomes:
1st outcome: Titans win all 4 games & lose 0
2nd outcome: Titans win 3 & lose 1
3rd outcome: Titans win 2 & lose 2
4th outcome: Titans win 1 & lose 3
5th outcome: Titans win 0 & lose 4
From Slick - 10/31/06 8:16 PM
14. ...answer a True/False test with 20 questions?
20 x 2 choices = 40
From Pringle - 10/31/06 10:36 AM
24...order a drink from an ice cream parlor that offers three drinks, sodas, milk shakes, and blasters in three sizes, small, medium, and large, with 24 flavors. If you order one drink, how many choices do you have?
24*3*3= 216 ( 24 flavors * 3 sizes * 3 drinks)
From wHolt - 10/30/06 11:00 AM
Can you check your thinking by listing all the ways your answer says exists?
Listing is the sure way of counting.
Both Kathi and Boki multiplied two numbers to get their answers.
Are they both correct?
It's important in these problems not to mutliply numbers together
without understanding what you are multiplying.
You need ways to check your thinking.
One way is to make your sets of objects smaller,
and test your formula by listing all possibilities in the smaller sets.
From Boki - 10/30/06 9:33 AM
11. ...walk from the center of a wild flower with 8 stems to the end of a flower petal, if each stem has 5 petals? (Pretend you are a bug living in the middle of the flower.)
With 8 stems and 5 petals on each of them I (as a bug) have 8*5 = 40 possibilities to walk from the center of a wild flower to the end of a flower petal.
From Kathi - 10/30/06 6:43 AM
23. ...watch a team win or lose the first four games of a World Series?
4*2 = 8
There are 4 games and either the team wins or loses (2 choices). 4 games * 2 choices = 8 outcomes.
From wHolt - 10/29/06 11:41 AM
| # | Alias | Date Started |
| 1 | ||
| 2 | ||
| 3 | ||
| 4 | ||
| 5 | ||
| 6 | ||
| 7 | ||
| 8 | ||
| 9 | ||
| 10 | ||
| 11 | Boki | 10/30 |
| 12 | DirtyBird | 11/06 |
| 13 | Harkar | 11/06 |
| 14 | Slick | 10/31 |
| 15 | Tiger | 11/10 |
| 16 | Poovey | 11/02 |
| 17 | Pac | 11/06 |
| 18 | JooJoo | 11/04 |
| 19 | SuperDuke | 11/05 |
| 20 | Centerfield | 11/12 |
| 21 | Sunshine | 11/12 |
| 22 | 7Iron | 11/10 |
| 23 | Kathi | 10/30 |
| 24 | Pringle | 10/31 |
| 25 | Taurus | 11/09 |
| 26 | Capricorn | 11/09 |
| 27 | David | 11/11 |
| 28 | Zonino | 11/13 |
| 29 | ||
| 30 | Draco | 11/13 |
| 31 | Fro | 11/16 |
| 32 | ||
| 33 | ||
| 34 | SkoolGirl | 11/13 |
| 35 | ||
| 36 | ||
| 37 | ||
| 38 | BassLady | 11/17 |
| 39 | Houdini | 11/12 |
| 40 |
From wHolt - 10/29/06 8:31 AM
Now that you have mastered everything there is to know about numbers,
you are ready to learn how to count things.
It's like moving from kindergarten to first grade.
For the next 5 assignments you will discover new ways to count.
Pay particular attention to this first assignment called Counting
because all the other ways of counting are based on the first way.
Counting became important when people started owning property.
They needed to assign a number to each item they owned, like sheep and coins.
Today, with all the many ways to arrange pipes and wires and switches,
we need to know how to count big numbers in a short time.
We don't want to count a billion things the long way 1, 2, 3,..., 1 billion.
We need rules for shortcuts.
As a group, you will discuss your ideas, and decide what these rules are.
This group of assignments will be worth 7 points each,
with each of you responsible for solving one problem.
Remember:
YOU MAY NOT USE ANY SYMBOLS IN YOUR SOLUTION EXCEPT DIGITS (0-9)
AND THE 4 OPERATORS + - * /
Last Modified 11/20/06 10:02 AM
even though reel one has only 1 Bell,
it should be represented in your Stem box.
You only represented reels 2 and 3.