If we now write down the table of symbols in which the (m + 1)th

0123&c.
10₁1₁2₁3₁,&c.
20₂1₂2₂3₂,&c.
30₃1₃2₃3₃,&c.
&c. &c.&c.&c.&c.

number of the nth row represents mₙ, 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:

012345678910
1 11000000000
2 12100000000
3 13310000000
4 14641000000
5 1510105100000
6 16152015610000
7 172135352171000
8 1828567056288100
9 1936841261268436910
10 1104512021025221012045101

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:

1 1   1 2 1   1 3 3 1
1 1 1 2 1 1 3 3 1
1 2 1 1 3 3 1 1 4 6 4 1, &c.

so that the sum in each row must be double of the sum in the preceding. But we can carry the consequences of this mode of formation further. If we make the powers of 1 + x by actual algebraical multiplication, we see that the process makes the same oblique addition in the formation of the numerical multipliers of the powers of x.

Here are the second and third powers of 1 + x: the fourth, we can tell beforehand from the table, must be 1 + 4x + 6x² + 4x³ + x⁴; and so on. Hence we have