0

I have a problem that goes like this:

At the beginning of every period of British Literature, Mrs. Crabapple picks a random student to receive a crabapple as a gift, but really, as you might imagine, they are quite bitter and nasty. Given that there are 11 students in her class and her class meets four times a week, how many different sequences of crabapple recipients are possible in a week, if no one student receives more than one apple a week?

I know the answer to this is to calculate 11*10*9*8 which equals 7920

What I want to know is, why does 11 choose 4 not work? According to what I know about combinations, we can calculate 11 choose 4 to find how many different ways can we pick 4 students from 11 to receive apples. Doesn't this properly answer the question that was asked? I feel that I have a fundamental misunderstanding of what combinations do, and would really appreciate if someone could help me understand what I am missing.

Thanks for your help :)

K. King
  • 129

2 Answers2

1

We are not interested in "which four students get crabapples", but rather "which four students get crabapples on which day". Those are two different things, although you can get from one to the other by multiplying or dividing by $24$.

Arthur
  • 199,419
1

11 choose four does give you the 4 students that can receive an apple. Let's say you chose students A,B,C and D

But in a week, the combination also takes inti account when they receive the gift. So you need to order them in the days they meet. So for example let ABCD denote the possibility that A was first, then B, then C, then D. Clearly it's not the same as DCBA. And for every 4 students you choose you get 4! possibilities in which you order them in the week (same as the permutations of the word ABCD)

So the answer should be$$ 4! {11 \choose 4} = 4! \times \frac {11!}{4!7!} = 11*10*9*8 = Var(11,4)$$

A variation denotes the choosing of elements in a set and then ordering them. You want to get a variation because the order matters in this case

For reference, $Var(n,r) = \frac{n!}{(n-r)!}$