Question

Hello everyone,
Having trouble with a problem:
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Mrs. Moyer’s class has to choose 4 out of 12 people to serve on an activity committee.
A) Suppose the students are selected for the positions of chairperson, activities planner, activity leader, and treasurer. How many different groups of students could be selected?
--> I say 12P4 = 12! / (12-4)! = 11,880
B) What is the probability that any one of the students is chosen to be the chairperson?
--> I say (1 * (11 * 10 * 9)) / 11,880 = 1/12
--> Book says 1/2970
Can't figure out if I am wrong or t

Answer 1

i think u just need to divide it by four

Answer 2

cause 11880 is the probability for all the job

Answer 3

Thanks for replying. :) I did notice that 4/11880 = 1/2970 but I'm not understanding it. I tried to imagine all 11880 outcomes with 990 of them starting with person 1, let's call him John. But even though it makes sense to me, I guess I'm wrong and just not getting it. Could you explain your thoughts a little more, please?

Answer 4

Well, she has to choose 4 people, so that means there are:
I would say you are correct with Question A, after all:
12! / (12 - 4)! = 11,880 (And unfortunately I can't think straight with the probabilities, but again I believe you're correct.)
Yeah I came out to 1/12 as well for Question B, hmmm let me check to see if I can figure out what it is we're missing.

Answer 5

Well, I can't find anything unfortunately, sorry that I couldn't help you out.
I hope you have a great rest of your day at least. :/
{---Ghostgate---}

Answer 6

Thanks for trying. :)

Answer 7

For A, you need the number of ways to arrange 12 things in groups of 4:
$$
{12\choose 4}=\frac{12!}{4!\left (12-4\right )!}=\frac{12!}{4!8!}=495
$$
For B, to be chosen chairperson you need be chosen to be in a group and then chosen to be chairperson. There are 495 ways of groups of 4 to be arranged from 12 people. Once chosen for a group, there are 4! ways for that group to be arranged and a single position will be chairperson for 1/4 of these possibilities, or 4!/4 = 3!

So total number of ways to be chosen chairperson is
$$
{12\choose 4}3!
$$
Finally, the probability that a person will be in a group and chosen as chairperson is the inverse of this:
$$
\frac{1}{{12\choose 4}3!}=\cfrac{1}{2970}
$$