There are curious and useful relations existing between numbers of combinations, some of which may readily be exhibited, under the simple expression of _m_ₙ to stand for the number of ways in which _m_ things may be taken out of _n_. Suppose we have to take 5 out of 12: Let the 12 things be marked A, B, C, &c. and set apart one of them, A. Every collection of 5 out of the 12 either does or does not include A. The number of the latter sort must be 5₁₁; the number of the former sort must be 4₁₁, since it is the number of ways in which the _other four_ can be chosen out of all but A. Consequently, 5₁₂ must be 5₁₁ + 4₁₁, and thus we prove in every case, _m_ₙ′ = _m_ₙ₋₁ + (_m_ - 1)ₙ₋₁ 0ₙ and _n_ₙ both are 1; for there is but one way of taking _none_, and but one way of taking _all_. And again _m_ₙ and (_n_-_m_)ₙ are the same things. And if _m_ be greater than _n_, _m_ₙ is 0; for there are no ways of doing it. We make one of our preceding results more symmetrical if we write it thus, 2ⁿ = 0ₙ + 1ₙ + 2ₙ + ... + _n_ₙ If we now write down the table of symbols in which the (_m_ + 1)th 0 1 2 3, &c. +-------------------------------------- 1 | 0₁ 1₁ 2₁ 3₁, &c. 2 | 0₂ 1₂ 2₂ 3₂, &c. 3 | 0₃ 1₃ 2₃ 3₃, &c. &c. | &c. &c. &c. &c. number of the _n_th row represents _m_ₙ, the number of combinations of _m_ out of _n_, we see it proved above that the law of formation of this table is as follows: Each number is to be the sum of the number above it and the number preceding the number above it. Now, the first row must be 1, 1, 0, 0, 0, &c. and the first column must be 1, 1, 1, 1, &c. so that we have a table of the following kind, which may be carried as far as we please: 0 1 2 3 4 5 6 7 8 9 10 +---------------------------------------------------- 1 | 1 1 0 0 0 0 0 0 0 0 0 2 | 1 2 1 0 0 0 0 0 0 0 0 3 | 1 3 3 1 0 0 0 0 0 0 0 4 | 1 4 6 4 1 0 0 0 0 0 0 5 | 1 5 10 10 5 1 0 0 0 0 0 6 | 1 6 15 20 15 6 1 0 0 0 0 7 | 1 7 21 35 35 21 7 1 0 0 0 8 | 1 8 28 56 70 56 28 8 1 0 0 9 | 1 9 36 84 126 126 84 36 9 1 0 10 | 1 10 45 120 210 252 210 120 45 10 1 Thus, in the row 9, under the column headed 4, we see 126, which is 9 × 8 × 7 × 6 ÷ (1 × 2 × 3 × 4), the number of ways in which 4 can be chosen out of 9, which we represent by 4-{9}. If we add the several rows, we have 1 + 1 or 2, 1 + 2 + 1 or 2², next 1 + 3 + 3 + 1 or 2³, &c. which verify a theorem already announced; and the law of formation shews us that the several columns are formed thus: