Question 1117539: Hello! I'm a little stumped on this problem. I'm not sure where to start. Any help is greatly appreciated and welcomed! It really helps me! Thank you so much!
The math club has 9 members
1. How many ways can we choose a statistics committee of 3 members?
2. How many ways can we choose a president, vice president, and secretary?
Thank you so much!
Found 2 solutions by ikleyn, solver91311:
Answer by ikleyn(52794)   (Show Source):
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
The way this question is worded makes me believe that the "Statistics Committee" is not ordered, whereas the three officers are ordered.
Let's tackle the second question first. You have nine members. There are then 9 ways to choose the President. Then, for each of the ways to choose the President, there are 8 ways to choose the Vice-President, for a total of 9 times 8 or 72 ways to choose the President and the Vice-President. Then, for each of the ways to choose the President and the Vice-President, there are 7 ways to choose the Secretary, 72 times 7 = 504. Note that in this calculation, the order of selection is important. President Mary, Vice-President Ann, and Secretary Bill is a different outcome than President Ann, Vice-President Bill, and Secretary Mary.
In general, the Permutation of  things taken  at a time is given by:
!})
Applying this to the problem of selecting club officers:
!}\ =\ \frac{9!}{6!}\ =\ \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}{6 \times 5 \times 4 \times 3 \times 2}\ =\ 9\ \times\ 8\ \times\ 7\ =\ 504)
Now we consider the question of the committee. We are still choosing 3 things out of 9 things, but this time, order doesn't matter. A committee of Mary, Ann, and Bill is the same as a committee of Ann, Bill, and Mary.
Note that, in a set of 3 people, there are three ways to choose the first person, two ways to choose the second person, and one way to choose the last person, 3 times 2 times 1 = 6 different orders.
That means that the first calculation we made has 6 instances of every possible set of 3 of the 9 members, and that means the answer to the question about officers is 6 times larger than the answer we need for the committee members.
In general, the number of Combinations of things taken at a time is given by:
!})
Let's apply this to the Committee problem
!}\ =\ \frac{9!}{6!}\ =\ \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}{(3 \times 2)(6 \times 5 \times 4 \times 3 \times 2)}\ =\ 3\ \times\ 4\ \times\ 7\ =\ 84)
Notice the extra factor of 6 in the denominator, the number of ways to order 3 things.
John
My calculator said it, I believe it, that settles it
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