Amalgams

#19.  Select 3 outfielders from 9 baseball players?

 I don't think anyone has chosen this problem yet.  To select 3 outfielders from 9 baseball players - 9 * 8 * 7 = 504 different ways

Pac - compute #17 when order does matter.
Which way gives the bigger number?
Should the number be bigger one way or the other?

Kathi - you are only choosing 2 letters, not 26.
How many ways to choose the first?
How many ways to choose the second?

26 * 1 was a short way to figure the answer.
 you have 26 choices and can only choose one of the 26.
which in turn means: a or b or c or d or e or  ......... or z
you can choose a which is one choice or b which would be a 2nd choice or c and so on.
1 + 1 + 1 + 1 + 1  + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 26 choces
or as I first put 26 * 1 = 26 choices

The last post was trying to answer your next question was this the wrong way to find the answer to the question?

if you were choosing 2 letters from the alphabet,
would you multiply 26 x 2 to count the ways of choosing 2 letters?
How many ways can you choose 2 letters  from the alphabet?

26x25x24x23x22x21x20x19x18x17x16x15x14x13x12x11x10x9x8x7x6x5x4x3x2x1=403291461126605635584000000
403291461126605635584000000/2x1 = 201645730563302817792000000
This would mean there are 201645730563302817792000000 ways to choose 2 letters from 26 letters in the alphabet.

In order to allow you to find the ways you can choose the two letters without repeating the same letter as the first you need to divide the number of ways you can choose two letters by the number of duplicate ways or 2x1.

n(n-1)(n-2)(n-3)...../2*1

This means there are 201645730563302817792000000 ways to choose two letters without repeating the same two.

From wHolt - 11/20/06 11:02 AM

How many ways can you choose 2 letters  from the alphabet.

26x25x24x23x22x21x20x19x18x17x16x15x14x13x12x11x10x9x8x7x6x5x4x3x2x1=403291461126605635584000000
403291461126605635584000000/2x1 = 201645730563302817792000000
This would mean there are 201645730563302817792000000 ways to choose 2 letters from 26 letters in the alphabet.

In order to allow you to find the ways you can choose the two letters without repeating the same letter as the first you need to divide the number of ways you can choose two letters by the number of duplicate ways or 2x1.

n(n-1)(n-2)(n-3)...../2*1

This means there are 201645730563302817792000000 ways to choose two letters without repeating the same two.

 

Part 1:
     A. Arrangements of K things out of N things: You multiply K number of items, starting with N, followed by (N-1), (N-2), and so on, until you fill K positions. For example:
     |  K  |  N  |     (N)*(N-1)*(N-2)...  
     |  2  |  10 |    10*9 = 90
     |  5  |  10 |    10*9*8*7*6 = 30,240
     | 10  |  10 |    10*9*8*7*6*5*4*3*2*1 = 3,628,800

     B. Subsets of K things out of N things: Follow the same pattern as mentioned above, but divide the product by (K)*(K-1)*(K-2)... for K number of positions.
     |  K  |  N  |     (N)*(N-1)*(N-2).../(K)*(K-1)*(K-2)... 
     |  2  |  10 |    10*9 = 90/(2*1) = 45
     |  5  |  10 |    10*9*8*7*6 = 30,240/(5*4*3*2*1) = 252
     | 10  |  10 |    10*9*8*7*6*5*4*3*2*1 = 3,628,800/(10*9*8*7*6*5*4*3*2*1) = 1

     C. How is your rule similar to the rule for counting arrangements of subsets? The rule in B is similar to that of A because both of them begin with the same pattern...

     D. How does your rule differ from the rule for counting arrangements of subsets? ...The difference is that the rule in B goes a step further and incorporates a division sequence onto the multiplication pattern established in A; the purpose is to eliminate duplicate answers since order does not matter when dealing with subsets (as it does matter when dealing with arrangements).

Part 2: How many ways can you...
     #16. Elect 3 people for a committee from a fraternity with 16 members?
     16*15*14 / 3*2*1 = 560

Not enough of you are telling me why you are dividing by some number.
Are you copying from a book?
Then why does the book tell you to divide?
Go beyond the formula...

Write your problem number or I others will steal your claim.

JooJoo - Instead of dividing by 8x7x6x5x4x3x2x1
write down how many ways to pick the
1st store
2nd store
3rd store
.
.
.
.
8th store

use these numbers to obtain your answer
and remember,
in this assignment the order of the stores does not matter
however, this problem has already been claimed.
pick another

Tiger - this assignment does not count the order

Pringle loses a point for using !

Nice diagrams from David.

n=4, k=3

A. 4x3x2x1/3x2x1=4
at the risk of using one of my own creations twice, and since I need to write everything out.....the number of letters on each of the groups last row represents the possible number of Arrangements of k things out of n things.

Since k=3 we can disregard the top row with only four letters on it. Each one of the four groups represents a single "subset of k things out of n things."

B.  n*(n-1)*(n-2)*(n-3)...*(n-k)/n-k*(n-k-1)*(n-k-2)...*(n-k-k)/k*(k-1)(k-2)...*(k-k)

C. This is similar because the first two functions,n(n-1)...(n-k), and,(n-k)(n-k-1)...(n-k-k), is the same as the amalgam formula. n(n-1)...(n-k) finds the total number of ways to arrange items.

D. This is different because in the other excercises, duplicate combinations didn't matter.

Assignement part dux

2. Select 2 different weeks in the year for a vacation?

Im sure WHolt will want a better explanation but its thanksgiving, turkey is ready.

52*51/2*1=1326 ways to pick two different weeks of the year for vacation.

a. How many k letter subsets can you make from n letters?

    n!/ (n-K)! / K!

b. Write a rule that counts the subsets of k things taken from n different things. 
Write a rule where n is any number of letters,
and k is the number of letters that you select from the n letters.

 count n letters n*(n-1)*(n-2)*(n-(n-1)) and then divided by (n-k) ! and then this has to be divided by k! to eliminate duplicated used numbers. 

c. How is your rule similar to the rule for counting arrangements of subsets? 
     fomula n!/ (n-k)! is used. 
d. How does your rule differ from the rule for counting arrangements of subsets?

     n!/ (n-k)! has to be divided by k! again. 

 Assignment 2 I picked # 11

 11. Pick 11 players out of 20 for a football team?

 20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1/9*8*7*6*5*4*3*2*1 

 =   6704425728000 / 11*10*9*8*7*6*5*4*3*2*1 = 167960

Part 1
a - n=2, k=2
There are 2 possible arrangements for the 1st letter and 1 arrangement for the second letter giving us 2x1 for the arrangements.

If we take two letters then we could have the following subsets:
a
b
ab
neither a or b.
The only one that has 2 items as a subset of n is ab, so there is 1 subset that has two letters in it.

b -  The rule would be to find the total # of arrangements by using the formula n*(n-1)*(n-2)...(n-k+1) to determine the total number of arrangements for k out of n things.  To remove the duplications, we would then divide by k(k-1)*(k-2).. Thus if we wanted to pick 2 out of 5 things, we would take 5*4 which is 20 and then divide by 2*1 which is 2 to result in 10 possibilities.

c - The rule is similar to the rule of finding arrangements in that we are trying to remove all duplicate arrangements from our count.  In this way we do not count the same subset twice.

d - The rule is different for counting arrangements because we are only trying to find k number of n arrangements, and not the full number of arrangements.  The formula is actually a combination of finding the number of k arrangements of n things and finding the number of distinguishable arrangements of n things.

Part 2

Decide which of 13 homes to deliver 3 Looloolo pizzas?

We are trying to determine how we can deliver 3 pizzas to 13 homes.  We would first find the number of ways to deliver 3 pizzas to 13 homes, which would be 13*12*11 per our formula above.  That gives us 1,716 ways.  To eliminate the duplicates, we would take 1,716 and divide by 3*2*1 to find the distinguishable arrangements.  This leads us to 1,716/6 which is 286 ways.

a. How many k letter subsets can you make from n letters?

n=6 k=6

6*5*4*3*2*1=720

720/720*1=1 subset

b. Write a rule that counts the subsets of k things taken from n different things. 
Write a rule where n is any number of letters,
and k is the number of letters that you select from the n letters.

Multiply n by each number less than n. Then divide that number by the product of k multiplied by every number less than k, and the product of n-k multiplied by every number less than n-k.

c. How is your rule similar to the rule for counting arrangements of subsets?

In both rules you must divide by n-k.

d. How does your rule differ from the rule for counting arrangements of subsets?

In this rule you must divide by k, and THEN mutiply the quotient by n-k.

Choose 10 state capitals in any order to visit?

10272278170000000000 ways to vistit any 10 of the 50 state capitals.

n=10*9*8*7*6*5*4*3*2*1=3628800

k=50*49*48*47...*1=3041409320000000000000000000000000000000000000000000000000000000000000000

n-k=

40*39*38*37...*1=815915283300000000000000000000000000000000000000000000000

3041409320000000000000000000000000000000000000000000000000000000000000000/3628800*815915283300000000000000000000000000000000000000000000000=

10272278170000000000

#12 How many ways can you eat a dozen (12) doughnuts from a box of 12.

12*11*10*9*8*7*6*5*4*3*2*1=479001600 ways to eat 12 doughnuts

A. How many k letter subsets can you make from n letters?

I am also going to use the number from my problem later in the assignment.

n=10 and k=8

10x9x8x7x6x5x4x3x2x1/8x7x6x5x4x3x2x1=90

B.  if n=10 and k=8

You would do nx(n-1)x(n-2)x(n-3).... and so on until at the end of the problem you have x1 then divide by doing the same with k so.... kx(k-1)x(k-2)x(k-3)... and so on until you get to x1 at the end.  then you divide the new n/new k and you get your answer

Ex: 10x9x8x7x6x5x4x3x2x1/8x7x6x5x4x3x2x1=90 when n=10 and k=8

Part II

#8. How would you shop 8 stores out of 10 in the mall?

10x9x8x7x6x5x4x3x2x1/8x7x6x5x4x3x2x1=90

So there is 90 ways to shop 8 stores out of 10 stores in the mall.