7. [6] There are 15 stones placed in a line. In how many ways can you mark 5 of these stones so that there are an odd number of stones between any two of the stones you marked?
Answer: 77 Number the stones 1 through 15 in order. We note that the condition is equivalent to stipulating that the stones have either all odd numbers or all even numbers. There are $\binom{8}{5}$ ways to choose 5 odd-numbered stones, and $\binom{7}{5}$ ways to choose all even-numbered stones, so the total number of ways to pick the stones is $\binom{8}{5}+\binom{7}{5}=77 .\left(\binom{n}{k}\right.$ is the number of ways to choose $k$ out of $n$ items. It equals $\left.\frac{n!}{k!(n-k)!}\right)$.