Problem 53 Description There are exactly ten ways of selecting three from five, 12345: 123, 124, 125, 134, 135, 145, 234, 235, 245, and 345 In combinatorics, we use the notation,$\displaystyle \binom 5 3 = 10$ . In general,$\displaystyle \binom n r = \dfrac{n!}{r!(n-r)!}$ , where $r \le n$, $n! = n \times (n-1) \times … \times 3 \times 2 \times 1$, and $0! = 1$. It is not until