MHB Is one form of the answer more right than the other

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The discussion revolves around selecting a committee of 12 from 10 men and 10 women, requiring an even number of women. One participant provided a detailed combinatorial solution, while the answer key used sigma notation for a more compact expression of the same problem. The key point of contention is whether one solution is more correct than the other, with emphasis on the clarity and efficiency of the sigma notation approach. It was noted that the initial answer might be missing a term, suggesting that completeness is crucial in combinatorial problems. Ultimately, both methods are valid, but the second approach offers a more streamlined representation.
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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 [math]{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}[/math]

The answer key gives

[math]\sum\limits_{i=1}^5 {10 \choose 12-2i}{10 \choose 2i}[/math]

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?
 
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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.
 
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