I have the following question:
How many ways are there to select $3$ candidates from $8$ equally
qualified recent graduates for openings in an accounting firm?
I was given the following solution to it: $\binom{8}{3}$, however, I got the answer: $8 \times 7 \times 6$ by doing the multiplication rule, I was wondering why my answer doesn't work.
My initial reason is that in the multiplication rule there is the underlying assumption that order matters and here obviously order doesn't matter. However, I am posting this here as that is the first time I come to such a realization and I am not sure that my reasoning is correct, especially since that I never learned multiplication rule with the idea of order in my mind. Any clarification would be appreciated.
Best Answer
Yes you are right. The formula for the 'choose' function is $$\frac{n!}{k!(n-k)!}$$ where k is the number of things you choose and n is the number of things you choose from. If you do $8 \times 7 \times 6$ you are working out the number of permutations (where order matters). But you're trying to work out the number of combination and therefore you need to exclude all combinations where the order is the same. E.g. abc is the same as acb = bac = bca = cab = cba
(n-k)! takes care of how many people to choose each time (aka 3). But k! allows you to neglect the order and make sure each combination is distinct from another.
If you did $\frac{8!}{8-3!}$ you'd get $8 \times 7 \times 6$ but that would only give you the number of permutations
You want combinations so you should do $\frac{8!}{3!8-3!}$
Hope this helps!