TITLE: Finding the cardinality of a set given a set property.
QUESTION [8 upvotes]: Suppose a set $B$ has the property that $|{X:X \in P(B), |X| = 6}| = 28$. Find $|B|$
My solution to this problem is as follows:
We know we have $28$ subsets of $B$ with $6$ elements each, so we can write $\frac{n!}{(n-6)!6!} = 28$.
We notice that $28=4*7$. We move $6!$ to the other side so now we have $(2*3*4*5*6)*(4*7) = \frac{n!}{(n-6)!}$. This means that $3*4*5*6*7*8 = n(n-1)(n-2)(n-3)(n-4)(n-5)$. It follows that $n = 8$, so $|B| = 8$.
Question: Is this solution correct and a good way to go about this problem? Is there another way to think about this problem?
Mention: This problem comes from The Book of Proof (Section 3.3 problem 4)

REPLY [4 votes]: Yes, you have correctly identified the equation $n!/6!(n-6)!=28$ is the equation to solve. Your approach seems quite efficient (I'm assuming that you realize that you must assume that $n\ge 0$ in the last step). 
Another approach would be to note that the equation requires $n\ge 6$ and note that $n!/6!(n-6)!$ is increasing with $n$ and then test with different $n$s. Once you reach $n=8$ you find a match and know that you can stop there as higher values of $n$ will result in larger LHS.