Answer, 5040.
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.
- 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 as, thus, ( ) bc ( ) ( ) d. If the as 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 as 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
- 3 × 2 × 1
or 120. 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.
- 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,