Question 1176391: A college club has 18 members, of which 10 are men and 8 are women. The manager of the club is one of the men. A committee of 5 members will be formed, where the administrator must be.
(a) How many committees can be formed that have 2 women and 3 men?
(b) How many committees can be formed in which there is no less than one woman?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
The underlined portion of "A committee of 5 members will be formed, where the administrator must be" tells us that the manager is on the committee.
So we're asking the equivalent question of "how many four person committees are possible if there are 2 men and 2 women?" since that 3rd male seat is taken by the manager (it's fixed in place and won't change).
We have 10-1 = 9 men to pick from and we have 2 seats to fill for them. On a committee, order doesn't matter because no one member outranks another. If order did matter, then there would be 9*8 = 72 permutations of men possible. However, order doesn't matter, so we cut that in half to get 72/2 = 36. We divide by 2 because a grouping like {A,B} is the same as {B,A}. The 72 double-counts the true count.
In short, we have 36 ways to pick the male portion of the committee. For the women, we have 8*7 = 56 permutations and 56/2 = 28 combinations. There are 28 ways to select the two women from the pool of 8 total.
Overall, there are 36*28 = 1008 ways to select the two men and two women to join the manager on the committee.
Answer: 1008
===================================================================================
Part (b)
"no less than one woman" is another way of saying "at least one woman" which is equivalent to "1 or more women".
Again the manager is already on the committee, which means we have 5-1 = 4 seats to fill.
We have 18-1 = 17 people to pick from. There are 4 seats to fill. Using the combination formula, there are 17 C 4 = 2380 different committees possible where we don't have any restriction on men or women.
We'll come back to this later. If you're curious how I got 17 C 4 = 2380, check out the section below.
Let's consider a committee of all men. This means there are 0 women on the committee.
The manager is in place so we have 4 seats to fill. There are 10-1 = 9 men and 4 seats, leading to 9 C 4 = 126 combinations.
Because we found there were 2380 different committees possible, and 126 of them are all men, this leads to 2380-126 = 2254 committees where there is at least one woman on the committee.
This is because the events "committee is all men" and "there is at least one woman on the committee" are complementary events. One or the other must happen. Both events cannot happen simultaneously.
Answer: 2254
===================================================================================
An example how the combination formula works.
Let's compute 17 C 4
We'll plug n = 17 and r = 4 into the combination formula below
n C r = (n!)/(r!*(n-r)!)
17 C 4 = (17!)/(4!*(17-4)!)
17 C 4 = (17!)/(4!*13!)
17 C 4 = (17*16*15*14*13!)/(4!*13!)
17 C 4 = (17*16*15*14)/(4!) ....... the "13!" terms cancel
17 C 4 = (17*16*15*14)/(4*3*2*1)
17 C 4 = (57120)/(24)
17 C 4 = 2380
|
|
|
