Is one form of the answer more right than the other

[find_the_fun]

A committee of 12 is to be selected from 10 men and 10 women. In how many ways can the selection be carried out if there must be an even number women?

I answered \(\displaystyle {10 \choose 2}{10 \choose 10}+{10 \choose 4}{10 \choose 8}+{10 \choose 6}{10 \choose 6}+{10 \choose 8}{10 \choose 4}+{10 \choose 10}{10 \choose 2}\)

The answer key gives

\(\displaystyle \sum\limits_{i=1}^5 {10 \choose 12-2i}{10 \choose 2i}\)

I can see how the two are equivalent but is one more correct than the other? I found the first answer just by thinking about it; was the second answer arrived at by applying a formula I'm unaware of? i.e. does the second answer imply a certain approach?

 

[MarkFL]

Re: is one form of the answer more right than the other

I have moved the topic, as this type of counting problem is the type typically encountered in an introductory statistics course.

Your answer is not quite correct. Can you spot the missing term?

Your answer key is taking advantage of sigma notation which can make writing sums much more compact.