# Thread: make a committee 4 men out of 6

1. ## make a committee 4 men out of 6

QUESTION 1]: To make a committee 4 men are to be chosen out of 6 candidates. What is the probability that certain two men will serve on that committee?

Folks, I don't need the answer. I'm learning Probability thus I need to think about it. I just need a hint.

I already have solved another problem similar to this. I'm not sure if it's correct or not.
It is:

QUESTION 2]: A committee of 3 must be chosen from 8 candidates. What is the probability that certain three people will be selected.

MY SOLUTION: I consider that 3 certain people a group. therefore we have 5+1 groups (5 individuals and 1 group of those certain people)

The result will be : C(1,6)/C(3,8)

Is this correct? If not, then I definitely can't solve QUESTION 1

Thank you in advance for your help

2. Originally Posted by Narek
QUESTION 2]: A committee of 3 must be chosen from 8 candidates. What is the probability that certain three people will be selected.
MY SOLUTION: I consider that 3 certain people a group. therefore we have 5+1 groups (5 individuals and 1 group of those certain people)
I have no idea what your thinking was there.
The committee has three members. There are three particular people.
So the is only one way to have all three on the committee.
$\displaystyle \dfrac{1}{\binom{8}{3}}=\dfrac{3!\cdot 5!}{8!}$

3. Therefore the answer to the first question will be:

(C(2,2) * C(2,4)) / C(4,6) = 2/5

right ?

4. Yes, it is correct,

BTW does you text material really use $\displaystyle C(k,N)$ for the combination of N taken k?

If so, how odd.