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 _n_th 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 _n_th 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 _n_th 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