Question 1188415: A committee of four is to be selected from a group of twenty people. How many different committees are possible, given the following conditions?
One person is the chair, one person is the secretary, and the other two are responsible for refreshment clean up.
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Let's focus on the first two slots for now. The order matters because the chairperson is different from the secretary.
We have 20 choices for the chairperson, then 19 choices for the secretary. That yields 20*19 = 380 permutations. Each time we arrive at a new seat, we count down by 1. Let A = 380 since we'll use it later.
Once we've filled those two slots, we have two additional seats where order does not matter. Both refreshment cleanup people have the same role/job.
We have 18 choices for the third slot and 17 choices for the fourth slot. That gives 18*17 = 306 permutations. However, order doesn't matter so we divide by 2 to get 306/2 = 153. This represents the number of ways to pick the remaining two spots on the committee. Let B = 153.
Overall, we have A*B = 306*153 = 58140 different ways to select the four person committee.
Answer: 58140
Answer by ikleyn(52787)  (Show Source):
You can put this solution on YOUR website! .
A committee of four is to be selected from a group of twenty people.
How many different committees are possible, given the following conditions?
One person is the chair, one person is the secretary,
and the other two are responsible for refreshment clean up.
~~~~~~~~~~~~~
There are 20*19 = 380 choices for the first two positions (380 ordered pairs).
And there are  = 9*17 = 153 choices for the pairs responsible for refreshments and cleaning up
(unordered pairs).
In all, there are 380*153 = 58140 different committees. ANSWER
Solved.
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