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.
- 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?1