Page:Elementary algebra (1896).djvu/341

PERMUTATIONS AND COMBINATIONS. 323 Note. If in formula (2) we put r = n, we have nCn = n n 0 = 1 0 but nCn = 1, so that if the formula is to be true for r = n, the symbol 0 must be considered as equivalent to 1.

Ex. From 12 books in how many ways can a selection of 5 be made, (1) when one specified book is always included, (2) when one specified book is always excluded?

(1) Since the specified book is to be included in every selection, we have only to choose 4 out of the remaining 11.

Hence the number of ways = 11C4 = 11 10 9 8 1 2 3 4 = 330.

(2) Since the specified book is always to be excluded, we have to select the 5 books out of the remaining 11.

Hence the number of ways = 11C5; 11x 10x9x8x7 1x2x8x4x5= 462.

395. The number of combinations of n things r at a time is equal to the number of combinations of n things n-r at a time.

In making all the possible combinations of n things, to each group of r things we select, there is left a corresponding group of n-r things; that is, the number of combinations of n things r at a time is the same as the number of combinations of n things n-r at a time;

nCr=nCn-r.

This result is frequently useful in enabling us to abridge arithmetical work.

Ex. Out of 14 men in how many ways can an eleven be chosen?

The required number = 14C11 = 11C3 = 14 13 12 1 2 3 = 364.

If we had made use of the formula 14C11, we should have had to reduce an expression whose numerator and denominator each contained 11 factors.

396. In the examples which follow it is important to notice that the formula for permutations should not be used until the suitable selections required by the question have been made.