Combination Problem

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February 18th 2006, 06:21 AM
jamesinsc

Combination Problem

I thought this would be a simple problem, but I can still not come up with the correct answer.
The problem is:
12 Students are in a class. Five can go to room A, Four to room B, and Three to room C. How many ways can this happen?

Since order is not important it is a combination problem and not a permuntation problem.

So by using nCr for {5,4} = 5 and {5,3} = 10 gives me a total of fifty combinations(using the fundamental counting principal).

Would I not just figure the combination of how many students can go into room B? nCr{4,1} = 4 and then multiply 5 x 10 x 4 ?

I get 200 for the answer but according to the quiz this is not the answer.

Could someone tell me what I am doing wrong?
Thanks in advance
James
February 18th 2006, 08:07 AM
CaptainBlack

Quote:

Originally Posted by jamesinsc

I thought this would be a simple problem, but I can still not come up with the correct answer.
The problem is:
12 Students are in a class. Five can go to room A, Four to room B, and Three to room C. How many ways can this happen?

Since order is not important it is a combination problem and not a permuntation problem.

So by using nCr for {5,4} = 5 and {5,3} = 10 gives me a total of fifty combinations(using the fundamental counting principal).

Would I not just figure the combination of how many students can go into room B? nCr{4,1} = 4 and then multiply 5 x 10 x 4 ?

I get 200 for the answer but according to the quiz this is not the answer.

Could someone tell me what I am doing wrong?
Thanks in advance
James

The first room can be filled in $12\times 11\times 10\times 9 \times 8=12!/7!$ ways (counting each
permutation as a distinct way of filling the room. Thus for each combination
we have counted 5! permutations, so the number of ways of filling the first
room is $12!/(7! 5!)$ .

Similaly for the next room except that we have only the $7$ left overs to use,
so there are now $7!/(3!4!)$ .

There are no options for who is allocated to room three once the first
two rooms are filled.

So our answer is:

$\frac{12!}{7!5!} \frac{7!}{3!4!}=\frac{12!}{5!4!3!}=27720 <br />$

RonL
February 18th 2006, 08:23 AM
jamesinsc

Thats it!!

Thanks Ron!!!
I was going about that one all wrong. It's hard for an old man to go back to school.
Thanks again
James
October 19th 2008, 12:43 AM
nameless virus

'tis maybe is an indistinguishable permutation

Quote:

Originally Posted by jamesinsc

I thought this would be a simple problem, but I can still not come up with the correct answer.
The problem is:
12 Students are in a class. Five can go to room A, Four to room B, and Three to room C. How many ways can this happen?

Since order is not important it is a combination problem and not a permuntation problem.

So by using nCr for {5,4} = 5 and {5,3} = 10 gives me a total of fifty combinations(using the fundamental counting principal).

Would I not just figure the combination of how many students can go into room B? nCr{4,1} = 4 and then multiply 5 x 10 x 4 ?

I get 200 for the answer but according to the quiz this is not the answer.

Could someone tell me what I am doing wrong?
Thanks in advance
James

(Sleepy).........'TIS COULD BE AN INDISTINGUISHABLE PERMUTATION

use the formula: n!/p!q!r!...

12!/5!4!3!
=479001600/17280
=27,720

MAYBE YOU COULD TRY THIS ONE...(Speechless)
October 19th 2008, 08:31 AM
Soroban

Hello, James!

Another approach . . .

Quote:

12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?

Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.

From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.

From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.

Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.