Combination problem

A

abhi

New member
Joined
Sep 17, 2012
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3
Hello,

This is my first post. Please correct me if this is not the right category.

I have tried using the normal formula I know to solve this combination problem, but doesn't give the right answer (which is supposed to be 431).

"A committee of 5 persons is to be formed out of 6 Americans and 5 Russians. In how many ways can this committee be formed if, in the committee, exactly 2 are Americans."
 
abhi said:
I have tried using the normal formula I know to solve this combination problem, but doesn't give the right answer (which is supposed to be 431).

"A committee of 5 persons is to be formed out of 6 Americans and 5 Russians. In how many ways can this committee be formed if, in the committee, exactly 2 are Americans."
Click to expand...

2 American out of 6 can be selected in 6C2 = 15 ways
3 Russians out of 5 can be selected in 5C3 = 10 ways

Committee={2A,3R}=15*10 = 150 ways.
 
MaxJasper said:
There is no such restriction in the site's guidelines.
Click to expand...
No, but the guidelines are meant for people asking questions, not those answering questions. Not giving answers is what most of us do most of the time because giving out answers does little to get students to understand what they are learning.

In this case, however, where the answer provided in the student's book is clearly wrong, I have more sympathy with giving out the correct answer. I'd still have phrased it that the answer given is wrong, what answer did you get.
 
Denis said:
431 certainly NOT correct... plus 431 is prime!

What did you get?

For fun: can you tell me what this represents (A1 = American#1...):
1 A1 A2 R1 R2 R3
2 A1 A3 R1 R2 R3
...
? A5 A6 R3 R4 R5
Click to expand...

Thanks for your response.

This I believe represents the number of possible combinations of the committee.

I tried the formula for combination - N!/(N-R)! R! - and got the answer as 150.
 
MaxJasper said:
2 American out of 6 can be selected in 6C2 = 15 ways
3 Russians out of 5 can be selected in 5C3 = 10 ways

Committee={2A,3R}=15*10 = 150 ways.
Click to expand...


Thanks for your response.

I got this answer as well, but I was not confident if this was correct.
 
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