This table exhibits the ways of making 1, 2, 3, and 4. Hence it follows (which I leave the reader to investigate) that there are twice as many ways of forming a + b as there are of forming a and then annexing to it a formation of b; four times as many ways of forming a + b + c as there are of annexing to a formation of a formations of b and of c; and so on. Also, in summing numbers which make up a + b, there are ways in which a is a rest, and ways in which it is not, and as many of one as of the other.
Required the number of ways in which a number can be compounded of odd numbers, different orders counting as different ways. If a be the number of ways in which n can be so made, and b the number of ways in which n + 1 can be made, then a + b must be the number of ways in which n + 2 can be made; for every way of making 12 out of odd numbers is either a way of making 10 with the last number increased by 2, or a way of making 11 with a 1 annexed. Thus, 1 + 5 + 3 + 3 gives 12, formed from 1 + 5 + 3 + 1 giving 10. But 1 + 9 + 1 + 1 is formed from 1 + 9 + 1 giving 11. Consequently, the number of ways of forming 12 is the sum of the number of ways of forming 10 and of forming 11. Now, 1 can only be formed in 1 way, and 2 can only be formed in 1 way; hence 3 can only be formed in 1 + 1 or 2 ways, 4 in only 1 + 2 or 3 ways. If we take the series 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, &c. in which each number is the sum of the two preceding, then the nth number of this set is the number of ways (orders counting) in which n can be formed of odd numbers. Thus, 10 can be formed in 55 ways, 11 in 89 ways, &c.
Shew that the number of ways in which mk can be made of numbers divisible by m (orders counting) is 2ᵏ⁻¹.
In the two series, 1 1 1 2 3 4 6 9 13 19 28, &c.
0 1 0 1 1 1 2 2 3 4 5, &c.,
the first has each new term after the third equal to the sum of the last and last but two; the second has each new term after the third equal to the sum of the last but one and last but two. Shew that the nth number in the first is the number of ways in which n can be made up of numbers which, divided by 3, leave a remainder 1; and that the nth number in the second is the number of ways in which n can be made up of numbers which, divided by 3, leave a remainder 2.
It is very easy to shew in how many ways a number can be made up of a given number of numbers, if different orders count as different ways. Suppose, for instance, we would know in how many ways 12 can be thus made of 7 numbers. If we write down 12 units, there are 11 intervals between unit and unit. There is no way of making 12 out of 7 numbers which does not answer to distributing 6 partition-marks in the intervals, 1 in each of 6, and collecting all the units which are not separated by partition-marks. Thus, 1 + 1 + 3 + 2 + 1 + 2 + 2, which is one way of making 12 out of 7 numbers, answers to
| 1 | 1 | 111 | 11 | 1 | 11 | 11 |
in which the partition-marks come in the 1st, 2d, 5th, 7th, 8th, and 10th of the 11 intervals. Consequently, to ask in how many ways 12 can be made of 7 numbers, is to ask in how many ways 6 partition-marks can be placed in 11 intervals; or, how many combinations or selections can be made of 6 out of 11. The answer is,
| 11 × 10 × 9 × 8 × 7 × 6 | , or 462. |
| 1 × 2 × 3 × 4 × 5 × 6 |
Let us denote by mₙ the number of ways in which m things can be taken out of n things, so that mₙ is the abbreviation for