A a time, is 3 5 times the number of combinations taken 3 at a time. Find n. 7. Out of 17 consonants and 5 vowels, how many words can be formed, each containing 2 vowels and 3 consonants? 8. Out of 12 consonants and 5 vowels how many words can be formed, each containing 6 consonants and 3 vowels? 9. The number of permutations of n things, 3 at a time, is 6 times the number of combinations, 4 at a time. Find n. 10. How many different sums may be formed with a guinea, a half-guinea, a crown, a half-crown, a shilling and a sixpence ? 11. At a game of cards, 3 being dealt to each person, any one can have 425 times as many hands as there are cards in the pack. How many cards are there? 12. There are 12 soldiers and 16 sailors. How many different parties of 6 can be made, each party consisting of 3 soldiers and 3 sailors? 13. If c2 be the total number of combinations of n things taken 1, 2, 3 ... n at a time, shew that 412. THE Binomial Theorem, first explained by Newton, is a method of raising a binomial expression to any power without going through the process of actual multiplication. 413. To investigate the Binomial Theorem for a positive integral index. By actual multiplication we can shew that (x+α1) (x+α2) = x2 + (α2+ α2) x +α12 (x+α1)(x+α2)(x+αz) = x2 + (α2+α2+αz) X2 (x+α2)(x+α2)(x+α3) (x+α ̧)=x2+ (α1 + A2 + A3 + A ̧) x3 + + In these results we observe the following laws: I. Each product is composed of a descending series of powers of x. The index of in the first term is the same as the number of factors, and the indices of a decrease by unity in each succeeding term. II. The number of terms is greater by 1 than the number of factors. III. The coefficient of the first term is unity. of the second the sum of a1, Ag, Ɑz ... of the third the sum of the products of a1, A2, A3 taken two at a time. ... of the fourth the sum of the products of a1, a2, az...taken three at a time. and the last term is the product of all the quantities A1, A2, A3.... Suppose now this law to hold for n-1 factors, so that (x+α1)(x+α2)(x+αz) (X +αn_1) ...... that is, the sum of the products of α1, A2, A3 ..... An−1, taken two at a time, that is, the sum of the products of a1, a2... an- 1 + Now S1+an=A2+ A2+ Az + ... + An-1+ an, that is, the sum of ɑ1, ɑ2, ɑ3 ..... Ans that is, the sum of the products of α1, ɑ2... ɑn, taken two at a time, S3+ An S2 = S3+ аn (α1a2+α2ɑ3+ ...), that is, the sum of the products of a1, a2... an, taken three at a time, anSn_1=α1α23... An_1 An, that is, the product of α1, ɑ1⁄2, ɑ3....... An• If then the law holds good for n-1 factors, it will hold good for n factors: and as we have shewn that it holds good up to 4 factors it will hold for 5 factors: and hence for 6 factors any number. and so on for за Now let each of the n quantities a1, a2, az an be equal to a, and let us write our result thus: -1 ... (x+α1) (x + α2) ... (x + αn) = x2 + A1. x2¬1 + Á2. x^−2+ The left-hand side becomes (x + a) (x+a) ... (x+a) to n factors, that is, (x+a)". And on the right-hand side A1=a+a+a+ ... A2 = a2+a2 + a2+ to n terms = na, ... to as many terms as are equal to the number of combinations of n things taken two at a time, 4th ... to as many terms as are equal to the number of combinations of n things taken three at a time, 414. Ex. Expand (x+a). Here the number of terms will be seven, and we have =x+6ax+15a3x+20a3x3 + 15a4x2+6u5x+a®. NOTE. The coefficients of terms equidistant from the end and from the beginning are the same. The general proof of this will be given in Art. 420. Hence in the Example just given when the coefficients of four terms had been found those of the other three might have been written down at once. 416. Every binomial may be reduced to such a form that the part to be expanded may have 1 for its first term. a2 + ... 1.2 |
