Page:Elementary algebra (1896).djvu/342

324 ALGEBRA. Ex. 1. From 7 Englishmen and 4 Americans a committee of 6 is to be formed: in how many ways can this be done, (1) when the committee contains exactly 2 Americans, (2) at least 2 Americans?

(1) The number of ways in which the Americans can be chosen is 4C2; and the number of ways in which the Englishmen can be chosen is 7C4. Each of the first groups can be associated with each of the second; hence

the required number of ways = 4C2 x 7C4 = 210.

(2) We exhaust all the suitable combinations by forming all the groups containing 2 Americans and 4 Englishmen; then 3 Americans and 3 Englishmen ; and lastly 4 Americans and 2 Englishmen.

The sum of the three results gives the answer. Hence the required number of ways = 4C2 x 7C + 4C3 x 7C3 + 4C4 x 7C2 = 210 + 140 + 21 = 371.

In this example we have only to make use of the suitable formula for combinations, for we are not concerned with the possible arrangements of the members of the committee among themselves.

Ex. 2. Out of 7 consonants and 4 vowels, how many words can be made each containing 3 consonants and 2 vowels ?

The number of ways of choosing the three consonants is 7C3, and the number of ways of choosing the two vowels is 4C2; and since each of the first groups can be associated with each of the second, the number of combined groups, each containing 3 consonants and 2 vowels, is 7C3 x 4C2.

Further, each of these groups contains 5 letters, which may be arranged among themselves in 5 ways. Hence the required number of words = 7 3 4 \times 4 2 2 \times 5 = 5 \times 7 = 25200.

EXAMPLES XXXV. a.

1. Find the value of 5P4, 7P6, 8C5, 25C23.

2. How many different arrangements can be made by taking (1) five, (2) all of the letters of the word soldier?

3. If nC3: n-1C4 = 8 : 5, find n.