SOLUTION: A committee of 5 people is to be chosen from a group of 6 men and 4 women. How many committees are possible if there are two men only?

Algebra ->  Permutations -> SOLUTION: A committee of 5 people is to be chosen from a group of 6 men and 4 women. How many committees are possible if there are two men only?      Log On


   

Question 1192407: A committee of 5 people is to be chosen from a group of 6 men and 4 women. How
many committees are possible if there are two men only?
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

There are n = 6 men and we select r = 2 of them.
Order doesn't matter on a committee since there aren't any special seats.
No member outranks another.

Order would matter if there were special labeled seats like president, chairman, VP, secretary, etc.

Because order doesn't matter, we use the nCr combination formula.
n C r = (n!)/(r!(n-r)!)
6 C 2 = (6!)/(2!*(6-2)!)
6 C 2 = (6!)/(2!*4!)
6 C 2 = (6*5*4!)/(2!*4!)
6 C 2 = (6*5)/(2!)
6 C 2 = (6*5)/(2*1)
6 C 2 = 30/2
6 C 2 = 15
There are 15 ways to select the two men.

If the 6 men have the code names A,B,C,D,E,F, then here are all of the fifteen combos of two men
  1. AB
  2. AC
  3. AD
  4. AE
  5. AF
  6. BC
  7. BD
  8. BE
  9. BF
  10. CD
  11. CE
  12. CF
  13. DE
  14. DF
  15. EF
The order doesn't matter so having a group like AB is the same as BA.

So far we've selected 2 positions for the committee and there are 5 total seats.
That leaves 5-2 = 3 seats for the women.
We have n = 4 women to pick from and r = 3 remaining seats to fill.

We can use the nCr combination formula again.
n C r = (n!)/(r!(n-r)!)
4 C 3 = (4!)/(3!*(4-3)!)
4 C 3 = (4!)/(3!*1!)
4 C 3 = (4*3!)/(3!*1!)
4 C 3 = 4/1
4 C 3 = 4
There are 4 ways to fill the three remaining seats with women.

This is equivalent to the fact that there are 4 ways to not select a particular unlucky woman who won't get on the committee.

If the women had code names P,Q,R,S, then here are all possible groups of three women
  1. QRS
  2. PRS
  3. PQS
  4. PQR
Like before, the order doesn't matter. A group like QRS is the same as RSQ.

Notice how,
group QRS leaves out P
group PRS leaves out Q
group PQS leaves out R
group PQR leaves out S


-----------------------------------

Let's wrap things up:
  • We found there are 15 ways to select the two men (from a pool of 6)
  • We found there are 4 ways to select the three women (from a pool of 4).
Multiply those results to get the final answer
15*4 = 60

Why multiplication? Because we can think of it like a table with 15 rows and 4 columns.
Each row/column combo represents a different configuration of the men and women groups combined.
This 15 by 4 table has 15*4 = 60 inner cells.

Answer: 60 different committees that have exactly two men.