Let us denote by _m_ₙ the number of ways in which _m_ things can be taken out of _n_ things, so that _m_ₙ is the abbreviation for _n_ - 1 _n_ - 2 _n_ - _m_ + 1 _n_ × ------ × ------- ... as far as ------------- 2 3 _m_ Then _m_ₙ also represents the number of ways in which _m_ + 1 numbers can be put together to make _n_ + 1. What we proved above is, that 6₁₁ is the number of ways in which we can put together 7 numbers to make 12. There will now be no difficulty in proving the following: 2ⁿ = 1 + 1ₙ + 2ₙ + 3ₙ ... + _n_ₙ In the preceding question, 0 did not enter into the list of numbers used. Thus, 3 + 1 + 0 + 0 was not considered as one of the ways of putting together four numbers to make 5. But let us now ask, what is the number of ways of putting together 7 numbers to make 12, allowing 0 to be in the list of numbers. There can be no more (nor fewer) ways of doing this than of putting 7 numbers together, among which 0 is _not_ included, to make 19. Take every way of making 12 (0 included), and put on 1 to each number, and we get a way of making 19 (0 not included). Take any way of making 19 (0 not included), and strike off 1 from each number, and we have one of the ways of making 12 (0 included). Accordingly, 6₁₈ is the number of ways of putting together 7 numbers (0 being allowed) to make 12. And (_m_- 1)ₙ₊ₘ₋₁ is the number of ways of putting together _m_ numbers to make _n_, 0 being included. 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_ - 2_b_ - 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.