If the hundredth part of a farthing be given for every different arrangement which can be made of fifteen persons, to how much will the whole amount? _Answer_, £13621608. Out of seventeen consonants and five vowels, how many words can be made, having two consonants and one vowel in each? _Answer_, 4080. 208. If two or more of the counters have the same letter upon them, the number of distinct permutations is less than that given by the last rule. Let there be _a_, _a_, _a_, _b_, _c_, _d_, and, for a moment, let us distinguish between the three as thus, _a_, _a′_, _a″_. Then, _abca′a″d_, and _a″bcaa′d_ are reckoned as distinct permutations in the rule, whereas they would not have been so, had it not been for the accents. To compute the number of distinct permutations, let us make one with _b_, _c_, and _d_, leaving places for the _a_s, thus, ( ) _bc_ ( ) ( ) _d_. If the _a_s had been distinguished as _a_, _a′_, _a″_, we might have made 3 × 2 × 1 distinct permutations, by filling up the vacant places in the above, all which six are the same when the _a_s are not distinguished. Hence, to deduce the number of permutations of _a_, _a_, _a_, _b_, _c_, _d_, from that of _aa′a″bcd_, we must divide the latter by 3 × 2 × 1, or 6, which gives 6 × 5 × 4 × 3 × 2 × 1 --------------------- or 120. 3 × 2 × 1 Similarly, the number of permutations of _aaaabbbcc_ is 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 --------------------------------- 4 × 3 × 2 × 1 × 3 × 2 × 1 × 2 × 1. EXERCISE. How many variations can be made of the order of the letters in the word antitrinitarian? _Answer_, 126126000. 209. From the number of permutations we can easily deduce the number of combinations. But, in order to form these combinations independently, we will shew a method similar to that in (206). If we know the combinations of two which can be made out of _a_, _b_, _c_, _d_, _e_, we can find the combinations of three, by writing successively at the end of each combination of two, the letters which come after the last contained in it. Thus, _ab_ gives _abc_, _abd_, _abe_; _ad_ gives _ade_ only. No combination of three can escape us if we proceed in this manner, provided only we know the combinations of two; for any given combination of three, as _acd_, will arise in the course of the process from _ac_, which, according to our rule, furnishes _acd_. Neither will any combination be repeated twice, for _acd_, if the rule be followed, can only arise from _ac_, since neither _ad_ nor _cd_ furnishes it. If we begin in this way to find the combinations of the five, _a_ _b_ _c_ _d_ _e_ _a_ gives _ab_ _ac_ _ad_ _ae_ _b_ ···· _bc_ _bd_ _be_ _c_ ···· _cd_ _ce_ _d_ ···· _de_