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 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 _x_(_x_ - 1)(_x_ - 2) ---------------------; 1 × 2 × 3 that of four is _x_(_x_ - 1)(_x_ - 2)(_x_ - 3) ------------------------------; and so on. 1 × 2 × 3 × 4 211. The rule may in half the cases be simplified, as follows. Out of ten counters, for every distinct selection of seven which is taken, a distinct combination of 3 is left. Hence, the number of combinations of seven is as many as that of three. We may, therefore, find the combinations of three instead of those of seven; and we must moreover expect, and may even assert, that the two formulæ for finding these two numbers of combinations are the same in result, though different in form. And so it proves; for the number of combinations of seven out of ten is 10 × 9 × 8 × 7 × 6 × 5 × 4 --------------------------, 1 × 2 × 3 × 4 × 5 × 6 × 7 in which the product 7 × 6 × 5 × 4 occurs in both terms, and therefore may be removed from both (108), leaving 10 × 9 × 8 ----------, 1 × 2 × 3 which is the number of combinations of three out of ten. The same may be shewn in other cases. EXERCISES.