206. We now proceed to find how many _permutations_, each containing one given number, can be made from the counters in another given number, six, for example. If we knew how to find all the permutations containing four counters, we might make those which contain five thus: Take any one which contains four, for example, _abcf_ in which _d_ and _e_ are omitted; write _d_ and _e_ successively at the end, which gives _abcfd_, _abcfe_, and repeat the same process with every other permutation of four; thus, _dabc_ gives _dabce_ and _dabcf_. No permutation of five can escape us if we proceed in this manner, provided only we know those of four; for any given permutation of five, as _dbfea_, will arise in the course of the process from _dbfe_, which, according to our rule, furnishes _dbfea_. Neither will any permutation be repeated twice, for _dbfea_, if the rule be followed, can only arise from the permutation _dbfe_. If we begin in this way to find the permutations of two out of the six, _a_ _b_ _c_ _d_ _e_ _f_ each of these gives five; thus, _a_ gives _ab_ _ac_ _ad_ _ae_ _af_ _b_ ... _ba_ _bc_ _bd_ _be_ _bf_ and the whole number is 6 × 5, or 30. Again, _ab_ gives _abc_ _abd_ _abe_ _abf_ _ac_ ... _acb_ _acd_ _ace_ _acf_ and here are 30, or 6 × 5 permutations of 2, each of which gives 4 permutations of 3; the whole number of the last is therefore 6 × 5 × 4, or 120. Again, _abc_ gives _abcd_ _abce_ _abcf_ _abd_ ... _abdc_ _abde_ _abdf_ and here are 120, or 6 × 5 × 4, permutations of three, each of which gives 3 permutations of four; the whole number of the last is therefore 6 × 5 × 4 × 3, or 360. In the same way, the number of permutations of 5 is 6 × 5 × 4 × 3 × 2, and the number of permutations of six, or the number of different ways in which the whole six can be arranged, is 6 × 5 × 4 × 3 × 2 × 1. The last two results are the same, which must be; for since a permutation of five only omits one, it can only furnish one permutation of six. If instead of six we choose any other number, _x_, the number of permutations of two will be _x_(_x_-1), that of three will be _x_(_x_-1)(_x_-2), that of four _x_(_x_ -1)(_x_-2)(_x_-3), the rule being: Multiply the whole number of counters by the next less number, and the result by the next less, and so on, until as many numbers have been multiplied together as there are to be counters in each permutation: the product will be the whole number of permutations of the sort required. Thus, out of 12 counters, permutations of four may be made to the number of 12 × 11 × 10 × 9, or 11880. EXERCISES. 207. In how many different ways can eight persons be arranged on eight seats? _Answer_, 40320. In how many ways can eight persons be seated at a round table, so that all shall not have the same neighbours in any two arrangements?[30] _Answer_, 5040. [30] The difference between this problem and the last is left to the ingenuity of the pupil.