Understanding the difference between combinations and permutations Question: There are $6$ men and seven woman in a club. A committee is to be formed. How many ways can we select a committee of five persons?
Answer: $C(13,5)$ or $\dfrac{13!}{8!5!}$
That is the answer given in the book, but my answer was originally $P(13,5)$, or $13\times12\times11\times10\times9$.
I thought my answer was right because there are $P(13,5)$ permutations to fit five people into the committee from $13$ total people, why is it a combination problem and not a permutation problem?
Thanks
 A: Yes, there are $P(13,5)$ ways to choose five people to line up in a certain order to form the committee. But you count each committee $5!$ times (for example {Amy, Bob, Carl, Doug, Ed} is the same as {Ed, Doug, Carl, Bob, Amy}, but you counted them as being different). So we need to divide by $5!$. By the way, I think you have a typo: the answer is $C(13,5) = 13!/(8!5!)$.
A: Permutations are used when one is concerned with order. For example, if you wanted to choose how many ways are there to arrange five people in a line, the answer will be different. In your case, a person first in line is the same as a person fourth in line, i.e., order does not matter. 
So combinations are used when the problem does not concern the ordering of objects, rather strictly taking combinations of objects. Hope this helps!
A: It is a combination problem because you don't care the order the people are selected.  A committee of ABCDE is the same as one of EBADC.  Note that $C(13,5)=\frac {13!}{5!(13-5)!}$ so your $3!$ in the denominator should be $5!$
A: When to use Permutations or Combinations
I've also illustrated the concept here. The basic difference is to understabd you have to just choose a community (i.e. selection) and not prepare a Seating chart for the selected committee (ordering)
Thus in this case the answer is C(13,5) is the way to select a committee of 5 people, while if the question was to prepare number of ways to seat 5 people in from 13 the answer would be P(13,5) or 5!xC(13,5).
