Wednesday, 16 May 2012

Four out of eight students will be selected for a $10000 scholarship. In how many ways can this be done?

Hello!
Although $10000 is a number, it does not influence the answer.

Let's select 4
students one by one.

The first student may be any of those 8 possible. After
this choice, one less student remain, 8 - 1 = 7. The second student may be one of those 7 that
remain (we cannot select the same student twice). 7 - 1 = 6 remain for at this step. The third
student may be one of those 6 that remain, 6 - 1 = 5 remain for at this step. Finally, the
fourth student may be one of those 5 that remain.

We have 8 possibilities at
the first step, 7 at the second step, 6 at the third, and 5 at the fourth. It seems that the
answer is 8 * 7 * 6 * 5 = 1680 ways. But it is not the case. Indeed, we can present 1680 lists
of 4 students, but some of them will be equivalent.

Because there are no any
specific roles between four students eligible for the scholarship, the list [1, 2, 3, 4] is
equivalent to, say, [2, 3, 1, 4]. In other words, the order of four chosen students is
insignificant.

How to make the correct answer from here? Note that each
4-tuple may be rearranged in 4 * 3 * 2 * 1 = 4! = 24 ways, and all these rearrangements are
equivalent. This means our number 8 * 7 * 6 * 5 is 4 * 3 * 2 * 1 times greater than the correct
answer.

So, the correct answer is (8 * 7 * 6 * 5) / (4 * 3 * 2
* 1).
Finish this calculation yourself. Also, read about number of combinations,
which can give you this answer directly.

href="https://www.mathsisfun.com/combinatorics/combinations-permutations.html">https://www.mathsisfun.com/combinatorics/combinations-per...

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