All the example problems followed a similar theme. You had to find the number of ways you can choose some \(k\) out of \(n\) items, if the order of choosing does not matter. We by now know the procedure to do so: First, pretend the order matters and pick \(k\) items, one item after the other, in \(n, n-1, n-2, \dots, n-k+1\) ways. To obtain the total number of ways to do that, we need to multiply them: \(n \times (n-1) \times \dots \times (n-k+1)\). Then, ask ourselves in how many ways can we order the items that we have chosen? Well, in \(k!\) ways, the number of permutations. Since the order does not matter, we need to divide the number found before by \(k!\).
The total number of combinations (choosing \(k\) out of \(n\) objects) is \[n \times (n-1) \times \dots \times (n-k+1)\div k! = \frac{n!}{k! \times (n-k)!}\] It is an important formula, and is often denoted with a special symbol \(\binom{n}{k}\), read “n choose k”. When solving the problems below, you can use the formula, or if you do not want to, just work them out individually, just like the examples!
Six girls – Ashley, Betty, Cindy, Donna, Eve and Fiona are members of a school maths circle (in another school obviously). In how many ways can you pick 4 of them to participate in a baths battle against the RGS team?