NUMBERS.
Hence is easily deduced a general rule of determining the number of com binations in any case whatsoever. Sup pose, for example the number of quan tities to be combined q, and the expo nent of combination n; the number of combinations will be qn qn 1 2 &c. till the number3 4 to be added be equal to n. Take q = 6 and n = 4, the number of combinations will be 6-4+ 1 6-4 + 2 6-4+3 1 2 3 6-4+4 6-3 6-2_6-1-6+0 4 1 2 3 4 3456 , 1234 If it be required to know all the possi ble combinations of the given quantities, beginning with the combinations of the several two's, then proceeding to three's, &c. we must add g +0,q 2 q-1 1 2 1 2 g-1 q +0, 8cc 3 1 2 3 4 Whence the number of all the possible combinations will 2 +1 2 q 2 g q 1 g 2 q---- 3 q q 1 3 +1 2 3 T 2 9 2 3 9 is the sum of 3 4 5 the unciz of the binomial raised to the power q, and abridged of the exponent of the power increased by unity q 1. Wherefore, since these uncim come out 1 + 1, by being raised to the power q; and since 1 + 1 is equal to 2, 2q will be the number of all the possible combinations. For example, if the num ber of quantities be 5, the number of pos sible combinations will be 25-6=32 6=26.
Prob. 2. Any number of quantities be. ing given, to find the number of all these changes which these quantities, combined in all the manners possible, can under go. Let there be two quantities a and b, their variations will be two ; consequent ly, as each of them may be combined with itself, to these there must be added two variations more. Therefore the num ber of the whole will be 2 -I- 2 = 4. If there were three quantities, and the ex. ponent of the variation 2, the combina tions will be 3, and the changes 3 ; to wit, a b, a c, b c, and b a, c a, c b ; to which if we add the three combinations of each quantity with itself a a, b b, c c, we shall have the number of changes 3-1- 3 ± 3 = 9.
In like manner, it is evident, if the given quantities were 4, and the exponent 2, that the number of combinations will be 6, and the number of changes like wise 6, and the number of combinations of each quantity with itself4 ; and there fore the number of changes 16; if with the same exponent the given quantities were 5, the number of changes would be 25 ; and in general, if the number of the quantities were n, the number of changes would be n'.
Suppose the quantities 3, and the ex ponent of variation 3, the number of changes is found 27 = 31,74z. a a a, a a b, a b a, b a a, a a e, a a, c a a, a b 6, b a b, bb a, a b c, b a c, b c a, ac b, c a b, c b a, acc,ca c,cca,bbb, bbc, cb b, bcb, b c c, c b c, c c b, c c c. In like man ner it will appear, if the quantities were 4, and the exponent 3, that the number of changes would be 64 = 41 ; and in general, if th e number of quantities was n, and the exponent 3, the number of changes would be n1.
By proceeding in this manner it will be found, if the number of quantities be n, and the exponent n, that the num ber of changes would be nn. Where fore, if all the antecedents be added, where the exponent is less, the number of all the possible changes will be found -1 ± &c. till the number subtracted from n leaves 1, because the beginning is from single quantities taken once.
Since, then, the number of all possible changes is in a geometrical progression, the first or smallest term of which is the largest an, and the denominator al ; it will be equal (a 1). Suppose n = 4, the number of all possi ble variations will be (41-4)χ(1-4)= 1020 3 Suppose again n = 24, the number of all the possible variations will be χ (24 1) = 52009658644. 406318986777955348250600 divided by 23 = 1391724288S8725299942512849340 2200. In so many various methods may the 24 letters of the alphabet be varied and combined among themselves.