There are 20 students in the high school glee club. They need to pick four students to serve as the student representatives for an upcoming trip. How many ways are there to choose the representatives?

a.) $4845$

b.) $6$

c.) $116280$

d.) $11480$

  • 3
    $\begingroup$ What are your thoughts ? $\endgroup$ – mwoua May 6 '13 at 20:25
  • 1
    $\begingroup$ Do you have any thoughts? Look up combinations in your text. $\endgroup$ – Ross Millikan May 6 '13 at 20:25

You are simply looking for the number of possible 4-combinations from the set of 20 students. To find out more about combinations, see here.

Most generally, choosing $k$ objects from a set of $n$ objects is given by

$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$

  • $\begingroup$ @Cortizol - thanks for the edit, that looks much nicer $\endgroup$ – SSumner May 6 '13 at 20:34

It's actually quite a simple problem and a perfect opportunity for some visualization.

Imagine the scenario: 20 students are standing in line. There are 4 seats. A person has to pick 4 students from the 20 to sit in those 4 seats.

First, he has 20 to pick from. After he picks one, he has 19 left to pick from, then 18, and then 17.

In all, the number of permutations is $20 \times 19 \times 18 \times 17$.

$$20 \times 19 \times 18 \times 17 = 116280$$

But since we're not looking for the number of permutations, but rather, the number of combinations, we have to adjust that value.

Since we had 4 seats with 4 students, the number of possible ways to have them seated is $4! = 24$, meaning there are 23 unneeded permutations for each permutation we computed.

In other words, the number of combinations:

$$\frac {116280}{4!} = 4845$$

Or as SSumner puts it:

$$\frac{n!}{k!(n - k)!} = \frac{20!}{4!(20 - 4)!} = \frac {20!}{24 \times 16!} = \frac {116280}{24} = 4845$$

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    $\begingroup$ if we have $4$ students($a$,$b$,$c$,$d$), and we have to choose $3$, then we have $\binom{4}{3}=4$ ways to do that ($abc$, $abd$,$acd$,$bcd$), not $4 \cdot 3 \cdot 2 = 24$. Maybe I am on some drugs... $\endgroup$ – Cortizol May 6 '13 at 20:58
  • $\begingroup$ Hmm. I think I'm the one on drugs. Give me a sec to get my head around this. $\endgroup$ – Mohamad Ali Baydoun May 6 '13 at 21:04
  • $\begingroup$ This is the correct number of permutations, not the correct number of combinations. You need to remove the cases where you select the same four students in different order $\endgroup$ – SSumner May 6 '13 at 21:06
  • $\begingroup$ Oh! I see my error. Thank you for pointing that out. $\endgroup$ – Mohamad Ali Baydoun May 6 '13 at 21:06
  • $\begingroup$ There we go. Sorry for that error. It just seemed so intuitive at first :P $\endgroup$ – Mohamad Ali Baydoun May 6 '13 at 21:13

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