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.
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 | ||||
| Of these, | ab | gives | abc | abd | abe | |
| ac | ···· | acd | ace | |||
| ad | ···· | ade | ||||
| bc | ···· | bcd | bce | |||
| bd | ···· | bde | ||||
| cd | ···· | cde | ||||
| ae be ce and de give none. | ||||||
| Of these, | abc | gives | abcd | abce | ||
| abd | ···· | abde | ||||
| acd | ···· | acde | ||||
| bcd | ···· | bcde | ||||
| Those which contain e give none, as before. | ||||||
Of the last, abcd gives abcde, and the others none, which is evidently true, since only one selection of five can be made out of five things.
210. The rule for calculating the number of combinations is derived directly from that for the number of permutations. Take 7 counters; then, since the number of permutations of two is 7 × 6, and since two permutations, ba and ab, are in any combination ab, the number of combinations is half that of the permutations, or (7 × 6)/2. Since the number of permutations of three is 7 × 6 × 5, and as each combination abc has 3 × 2 × 1 permutations, the number of combinations of three is
| 7 × 6 × 5 | . |
| 1 × 2 × 3 |