# Basic Probability using combinations [closed]

1. (a) A committee of 5 people is to be chosen from a group of 10 (6 men and 4 women) (i) How many committees of 5 members can be chosen from 10 people? (ii) How many of the committees from (a) will have exactly 2 women on them? a.i) = 252 What is a.ii) and how?

## closed as off-topic by Graham Kemp, JimmyK4542, Harish Chandra Rajpoot, Ken, DanielJul 28 '15 at 4:31

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• Welcome to Math.SE. This is a site where you can certainly post questions, but you have to show your attempt in solving them. Homework questions without self-efforts are also strongly discouraged. – Shailesh Jul 28 '15 at 3:01

There are ${^{10}\mathrm C_{5}}$ ways to choose 5 of 10 people. That equals $252$.

$${^{10}\mathrm C_{5}} = \dfrac{10!}{5!\,5!} = 252$$

You want to count the ways to choose 2 of 4 women and 3 of 6 men.

$${^{4}\mathrm C_{2}}\cdot{^{6}\mathrm C_{3}} = \dfrac{4!}{2!\,2!}\cdot\dfrac{6!}{3!\,3!} = 120$$

Can you not fill in the boxes?(yes; filled)

Thank you, I can fill in those boxes and that does yield the correct answer, but could you please explain why it is the 4C2 multiplied by the 6C3? – Lara 6 mins ago - Lara

@Lara You are selecting a committee of $5$ people out of $2$ of $4$ women and $3$ of $6$ men.   There are $^4 \mathrm C _2$ ways to select the women and, for each of these, there are $^6 \mathrm C _3$ ways to select the men.   So to count the total of ways to perform both these (sequential) tasks we must multiply.

• Thank you, I can fill in those boxes and that does yield the correct answer, but could you please explain why it is the 4C2 multiplied by the 6C3? – Lara Jul 28 '15 at 3:19
• @Lara You are selecting a committee of 2 of 4 women and 3 of 6 men (a committee of size 5). There are $^4C_2$ ways to select the women and, for each of these, there are $^6C_3$ ways to select the men. So to count the total of ways to perform both these (sequential) tasks we must multiply. – Graham Kemp Jul 28 '15 at 3:25