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Combinations

Consider now the problem of placing three balls, all of them colored red, in 10 boxes that are numbered 1, 2, 3, ….., 10. We want to know the number of ways the balls can be placed, if each box can hold only one ball. The answer is,

(10 × 9 × 8)/3!

In general, the number of ways of placing r balls of the same color in n numbered boxes is

[m(n – 1) (n – 2) … (n – r + 1)]/r! = n!/r!(n – r)!

The quantity n!/r!(n – r)! is also denoted C(n, r).

Let us consider an example:

Suppose a housekeeper wants to schedule spaghetti dinners three times each week. Imagine the spaghetti dinners as three balls and the seven days in the week as seven boxes; then the number of ways of scheduling is,

7!/(3!4!) = 35