If (2n)!/(2n−2)! = 132, what is the value of n?

Correct answer: 6

Use the factorial identity m!/(m−2)!=m(m−1). With m=2n, the given equation becomes (2n)(2n−1)=132. We need two consecutive positive factors whose product is 132. Since 12×11=132 and the even factor must be 2n, we get 2n=12, hence n=6. Direct verification gives 12!/10!=12×11=132. Therefore option B is correct. If n=5, the product is 10×9=90; if n=7, it is 14×13=182; and if n=8, it is 16×15=240. Thus none of the other options satisfies the original equation. The factorial cancellation is valid when the factorial arguments are defined.