Need help solving the following. I cannot find enough information in the text to begin solving the problem.
1/4 of the five element subsets of (1,2,3...n) contain the element 7, determine n. n >5.
There are $\displaystyle \begin{pmatrix}n \\ 5\end{pmatrix}= \frac{n!}{5! (n-5)!}$ subsets of {1, 2, 3, ..., n} that contain 5 numbers. 1/4 of that would be $\displaystyle \frac{n!}{5!(n-5)! 4}$ and "1/4 of the five element subsets of (1,2,3...n) contain the element 7" implies that must be an integer.
Yes, there is enough information given to solve for $\displaystyle n$.
HallsofIvy has told you that 1/4 of the number of subsets containing 5 elements is:
$\displaystyle N=\frac{n!}{4\cdot5!(n-5)!}$
Now, to find the number of subsets of cardinality 5 containing 7 as an element, we may use the fundamental counting principle to state:
$\displaystyle N={n-1 \choose 4}=\frac{(n-1)!}{4!((n-1)-4)!}=\frac{(n-1)!}{4!(n-5)!}$
Hence, we have:
$\displaystyle \frac{n!}{4\cdot5!(n-5)!}=\frac{(n-1)!}{4!(n-5)!}$
Now, you just need to solve for $\displaystyle n$.