This last amounts to the solution of the following: In how many ways can n counters (undistinguishable from each other) be distributed into m boxes? And the following will now be easily proved: The number of ways of distributing c undistinguishable
counters into b boxes is (b - 1)b + c - 1, if any box or boxes may be left empty. But if there must be 1 at least in each box, the number of ways is (b - 1)c - 1; if there must be 2 at least in each box, it is (b - 1)c- b-1; if there must be 3 at least in each box, it is (b - 1)c - 2b - 1; and so on.
The number of ways in which m odd numbers can be put together to make n, is the same as the number of ways in which m even numbers (0 included) can be put together to make n-m; and this is the number of ways in which m numbers (odd or even, 0 included) can be put together to make ½(n-m). Accordingly, the number of ways in which m odd numbers can be put together to make n is the same as the number of combinations of m-1 things out of ½(n-m) + m-1, or ½(n + m)-1. Unless n and m be both even or both odd, the problem is evidently impossible.
There are curious and useful relations existing between numbers of combinations, some of which may readily be exhibited, under the simple expression of mₙ to stand for the number of ways in which m things may be taken out of n. Suppose we have to take 5