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From an inspection of the above, we deduce the following

RULE. Add the number denoting the class of the combinations to the number of individual things, and from the sum subtract one, multiply this remainder continually by the successive decreas. ing terms of the series of natural numbers, until we reach the term denoting the number of individual things; then divide this product by the number of permutations, found by Case I., Art. 88, of a number of individual things denoted by the class of the combinations.

Examples. 1. How many combinations of the 5th class can be formed out of 10 individual things ?

SOLUTION. The number denoting the class added to the number of individual things gives, 10+5=15, from which, subtracting one,

14 x 13 x12x11x10—2002, is the no sentim we get 14. Therefore, 14 x 13 x 12x11x10_

enes 1 X 2 X 3 X 4 X 5 number of combinations sought."

2. How many different numbers, of four places of figures each, can be formed out of the nine digits ?

12x11x10 X 9=495.

1x 2x 3x4 3. How many different combinations, of six things at a time, may be formed out of 11 individual things ?

Ans. 8008. CASE II. If we form combinations, without repetitions, we shall have as follows:

COMBINATIONS WITHOUT REPETITIONS.

1) Comb. (a, b, c, d, e, f, g, h, i)

(Second Class.)

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