SECTION IX. ON PERMUTATIONS AND COMBINATIONS.

		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.

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

Also, since any combination of four, abcd, contains 4 × 3 × 2 × 1 permutations, the number of combinations of four is

7 × 6 × 5 × 4	,
1 × 2 × 3 × 4

and so on. The rule is: To find the number of combinations, each containing n counters, divide the corresponding number of

permutations by the product of 1, 2, 3, &c. up to n. If x be the whole number, the number of combinations of two is

x ( x - 1)	;
1 × 2

that of three is

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