That is, the number of combinations of n things taken (n-s) at a time is the same as the number of combinations of n things taken s at a time.

Examples.

1. In how many ways may a guard of 5 men be selected from a troop of 20 men?

Ans. 15504.

2. How many different eight-oared crews can be selected from 16 men? Ans. 12,870, positions undistinguished. 3. How many different parties of 5 each can be selected from 15 men?

4. If n+2C=11,C2, find n and check.

Ans. 3003.

5. If nCs=3nC's-1, find n in terms of s and check.
6. If 5nC=12nC1, find n and check.
7. If 5,C,=18,Cs, find s and check.

Ans. n=13.
Ans. s=4.

52. An Important Principle.-It is important to note carefully the principle used several times already, that if one thing can be done in a ways, and, without interfering with the first performance, another thing can be done in b ways, both things can be done in ab ways.

For instance, if a troop of 50 men and a captain are to be chosen from 55 men and 5 officers, since the men can be chosen in 55C50 ways, and the captain in 5, the troop can be made up in 5X55C50=17,393,805 ways.

Examples.

1. In how many ways may 5 articles be put into 3 boxes, each capable of holding all 5 articles?

Ans. 243. 2. How many signals may be made with 10 numeral flags, each signal to consist of from 1 to 3 flags, repetitions being allowed? Ans. 1110.

3. In how many ways can the crew of an eight-oared shell be seated if 4 of them can row only on the port side? Ans. 576. 4. A pitcher throws 4 different curves and a straight ball; in how many ways can he pitch 4 times? In how many if he throws differently each time? Ans. 625, 120. 5. How many baseball teams can be formed from 15 players, of whom 2 can pitch only and 2 can catch only? Ans. 1320.

6. How many four-letter arrangements can be made using the letters causing without repetitions, if consonants and vowels alternate? Ans. 144.

7. Of a crew of 8 men, 3 can row only on the port side, 2 only on the starboard, the rest on either. How many ways are there of seating the crew? Ans. 1728.

53. Product of Linear Factors.-In order to find the rule by which we can write down without multiplication the product of any number of factors (x+a), (x+b), (x+c), (x+d), etc., we will first perform the actual multiplication for a few factors:

cx2+(ac+bc)x+abc

x2+(a+b+c) x2+(ab+ac+bc)x+abc

Multiplying this result by (x+d), we get

x2+(a+b+c+d) x3 + (ab+ac+ad+be+bd+cd) x2

and so on.

+(bcd+acd+abd+abc)x+abcd,

It soon becomes evident, as we continue to multiply new factors into the product, that the following general rule is true:

The product of n factors (x+a), (x+b), (x+c), (x+d), etc., is a polynomial of degree n, consisting of (n+1) terms, in which the coefficient of the various powers of x are as follows: Of an

Unity;

Of n-1: The sum of all the numbers a, b, c, d, etc.;

Of xn-2:

The sum of the products of the numbers a, b, c, d, etc.,

taken two at a time;

Of xn-3: The sum of the products of the numbers a, b, c, d, etc., taken three at a time;

Of - The sum of the products of the numbers a, b, c, d, etc., taken four at a time;

and in general:

The coefficient of xn-k is the sum of the products of the numbers a, b, c, d, etc., taken k at a time.

The last term comes under this general formula, being the coefficient of 1, or of an-n, and equal to the product of all n of the numbers a, b, c, d, etc.

54. The Binomial Theorem.-If in the preceding theorem, the numbers a, b, c, d, etc., are all equal, the product of the n factors becomes (x+a)"; each product of two of the numbers a, b, c, d, etc., becomes a2; each product of three becomes a3, and so on; each product of k of the numbers becomes a*.

Since nC=nCnk, the coefficients at equal distances from the two ends of the expansion are alike. The expansion with coeffi

Note that the general term of (x+a)",

n(n−1) (n−2). ... (n−k+1)

is the (k+1)-th term.

k

akxn-k,

This expression for the general term can be applied to finding any required term of a particular expansion; e. g., to find the fourth term of (x+a) 12. Here k=3, n=12; and the required fourth term is

n(n-1) (n-2)

3

akon-k-
=

12.11.10
1.2.3

a3x=220a3xo.

When it is desired to write out a complete expansion, it may be noticed that any term of (x+a)" may be got by multiplying the preceding term by its exponent of x, dividing by one more than its exponent of a, lowering the exponent of x one unit and raising the exponent of a one unit. For instance,

The expansion of (x-a)" is given by the same theorem, since (x-a)" = (x+(-a))"; the only difference is in the signs, which are alternately + and -.

The letters x and a may of course stand for any values what

ever.

Examples.

1. Write the first five terms of the expansion of each of the following: (a+x)", (a-x)", (x-a)", (a+x)°.

2. Write the full expansion of each of the following: (x+a)3, (x−a)3, (a−x)®, (a+x)*.

3. Write the full expansions of the following:

5

(x+2)*, (1− 2)°, (3x+3a)o, (1+fx)*.

4. Find the seventh term of (a-x)10.

5. Find the fourth term of
of(x-
of (x — 2a)20. Ans.

3

6. Find the term containing 2 in

55. The Binomial Theorem for Fractional and Negative Exponents. It is proved by the Calculus that the use of the Binomial Theorem can be extended to fractional and negative powers of a binomial under certain restrictions.

gives an unending succession of terms if n is fractional or negative; but if x is numerically less than a (and only then), the sum formed by adding more and more of these terms will approach as a limit the value of (a+x)".

In such an application of the Binomial Theorem, it is convenient to use the following expansion, which holds true for any value of x numerically less than 1:

Ꮖ

For the sake of compactness and accuracy, the work is best arranged in the form shown in the following example: