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285. We know from Art. 277 that if any expression containing a be expanded in two different convergent series arranged according to ascending powers of x, the coefficients of like powers of a in the two series will be equal. By means of this very important principle many theorems can be proved.

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to the coefficient of xo in (- 1)"x", which is zero. [See also Art.

247, Ex. 3.]

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[Equate coefficients of x2 in (1 − x) ̄a × (1 − x)−2 and in (1 − x)—§.]

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The terms on the right which give r3n+1 are

x3n+1 (1 − x)3n+1 + x3n (1 − x)3n + x3n−1 (1 − x)3n−1 +
coefficient of x3n+1 will be found to be

; and hence the

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286. Expansion of Multinomials. Any multinomial expression can be expanded by means of the binomial theorem.

Since (p+qx + ræ2 +.....)” may be written in the form

9 x + x2+.....)", it is only necessary to consider

p" (1 +
expressions in which the first term is unity.

Ρ Р

n

Now in the expansion of {1 + ax + bx2 + cœ3 +...}' that is of {1+ (ax + bx2 + cx3 +...)}", by the binomial theorem, the general term is

n (n − 1) (n − 2).....(n − r + 1)

r

(ax + bx2+cx3 +...)";

also in the expansion of (ax + bx2 + cx3 +.....)”, r being a positive integer, the general term is by Art. 258

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where each of a, ß, y,... is zero or a positive integer, and a + B + y +... = r.

Hence the general term of the expansion of the multinomial is

n (n − 1) (n − 2).....(n − r + 1)

a By...

aabßcy...xa+2B+3y+....

To find the coefficient of any particular power of x, say of x, we must therefore find all the different sets of positive integral values (including zero) of a, B, y,... which satisfy the equation a + 2ß + 3y +... = k; the corresponding value of r is then given by r = a + B + y +....., and the corresponding coefficient is found by substituting

in the formula for the general term. The required coefficient will then be the sum of the coefficients corresponding to each set of values of a, ß, y....

Ex. 1. Find the coefficient of x5 in (1 − x + 2x2 – 3x3)−3 ̧

The values of a, ß, y which satisfy a +28+3y=5 will be found to be 0, 1, 1; 2, 0, 1; 1, 2, 0; 3, 1, 0; and 5, 0, 0. The corresponding values of r will be 2, 3, 3, 4 and 5 respectively; and the corresponding coefficients will be

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287. From the above example it will be seen that the process of finding even the first six terms in the expansion of a multinomial is very laborious; in many cases, however, the work can be much shortened, as in the following examples.

Ex. 2. Find the coefficient of x13 in the expansion of

(1+x+x2+x3+x4)−2.

- x5 -2

We have (1+x+x2 + x3 + x4)−2 =

-X

= (1 − x)2 (1 − x5)−2

= (1 - 2x + x2) (1 + 2x5 + 3x1o + 4x15 + ...).

Hence the coefficient required is zero.

Ex. 3. Find the coefficient of x" in the expansion of (1 + x + x2 + x3)−1.

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Hence the coefficient of x4r is 1, the coefficient of x1r+1 is −1, the coefficient of x4+2 is zero, and the coefficient of xr+3 is zero.

Thus the coefficient of x" is 1 when n is of the form 4r, it is 1 when n is of the form 4r+1, and it is zero when n is of either of the forms 4r+2 or 4r +3.

Ex. 4. Find the coefficient of x" in the expansion of (1+2x+3x2+4x3 + to infinity)".

Since 1+ 2x + 3x2 +

......

= (1 − x)-2, the required expansion is that of (1 - x)-2n; the coefficient of x" is therefore

2n (2n+1)... (2n +r− 1)

r

288. Combinations with repetitions. The number of combinations of n things a together of which p are of one kind, q of a second, r of a third, and so on, can be found in the following manner.

Let the different things be represented by the letters a, b, c,.......; and consider the continued product

(1+ax+a2x2+...+a2x2) (1+bx+...+b2x2) (1+cx+...+c*x′′)........

It is clear that all the terms in the continued product are of the same degree in the letters a, b, c,... as in x; it is also clear that the coefficient of x is the sum of all the different ways of taking a of the letters a, b, c,... with the restriction that there are to be not more than p a's, not more than q b's, &c.; so that the coefficient of x in the continued product gives the actual combinations required. Hence the number of the combinations will be given by putting a=b=c=... 1. Thus the number of the combinations of the n things a together is the coefficient of xa in

=

(1 + x + x2 + + x2) (1+x+

...

...

+ x2) (1+x+ + x*)...

...

Ex. 1. Find the number of combinations 7 together of 5 a's, 4 b's and 2 c's.

The number required is the coefficient of x7 in (1 + x + ... +25) (1 + x + ... x1) (1 + x + x2), that is in (1-x) (1 − x3) (1 − x3) (1 − x)-3. Rejecting terms of higher than the seventh degree in the continued

product of the first three factors, we have (1 − x3 – x5 — x6) (1 + 3x +6x2+10x3 + 15×a + 21x5 + 28x6 + 36x7+...); and the coefficient of x7 is 36-15-6-3=12.

Ex. 2. Find the total number of ways in which a selection can be made from n things of which p are alike of one kind, q alike of a second kind, and so on.

The total number of the combinations is the sum of the coef

...

ficients of x1, x2,...x" in (1 + x + ... x2) (1 + x + +x)...; and this sum is obtained by putting x=1 in the product and subtracting 1 for the coefficient of x°. Hence the required number is

(p + 1) (q+1)... – 1.

The above result can, however, be obtained at once from the consideration that there are p+1 ways of selecting from the a's, namely by taking 0, or 1, or 2,... or p of them; and, when this is done, there are q+1 ways of selecting from the b's; and so on.

Hence the total number of ways, excluding the case in which no letter at all is selected, is (p+1) (g + 1)... − 1. [Whitworth's Choice and Chance, Prop. XIII.]

Ex. 3. A candidate is examined in three papers to each of which m marks are assigned as a maximum. His total in the three papers is 2m; shew that there are (m + 1) (m + 2) ways in which this may

occur.

1 2

The number of ways is the coefficient of x2m in (1+x+x2+ ... xm)3, that is in (1 - xm+1)3 (1 − x)−3 — (1 − 3xm+1+ ...) ×

{1.2+2.3x+. +m (m + 1) xm−1 + ... + (2m + 1) (2m+2) x2m + ...}. Hence the number required

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{(2m+1) (2m+ 2) − 3m (m+ 1)} = — (m + 1) (m + 2).

289. Homogeneous Products. We have already [Art. 246] found the number of homogeneous products of r dimensions which can be formed with n letters, where each letter may be repeated any number of times. We now give another method of obtaining the result. Suppose the letters to be a, b, c,...; then if the continued product

(1 + ax + a2x2 + a3x3 +.....) × (1 + bx + b2x2 + b3ï3 +.....)

× (1+ cx + c2x2 + c3ï3 +.....)...

be formed, the coefficient of x" will clearly be of r dimensions in the letters a, b, c,..., and will be the sum of all the

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