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The first form is best adapted for ascending series; the second is derived from it by multiplying numerator and denominator by 1, and is best adapted for descending series.

79. If r be less than unity, the quantity" becomes less and less as n increases; and by taking n large enough, it may be made less than any quantity that can be assigned. Hence, if n be taken infinite, " becomes = 0, and therefore the formula

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From which it appears that to find the sum of an infinite number of terms of a descending geometrical series, we have only to divide the first term by the excess of unity above the common ratio. However great a number of terms of the series be taken, their sum will never actually amount to

but may be made to approach as near to it as we please.

EXERCISES.

Find the sum of each of the following series:

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(2.) 4, 3, 2, &c., to 10 terms.

(3.) 1, 1, 1, &c., to infinity.

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(4.) 1, — 1, 1, &c., to 9 terms and to infinity.

(5.) 333, &c., to infinity.

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૩.

A.

26 111.

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4+ 3/2.

(9.) The first term of an infinite geometric series is 1, and each term is equal to the sum of all that follow it; find the series. Ans. 1,,, &c.

(10.) The sum of an infinite geometric series is, and the sum of the first two terms is ; find the series.

6 18

Ans.,,, &c.; or 1, 1, 3, &c.

(11.) There are 4 numbers in arithmetical progression, which, being increased respectively by 1, 1, 2, and 41, are in geometrical progression; find the numbers.

Ans. 3, 5, 7, 9.

(12.) Find the nth term, and the sum of n terms of the series a+a, 2a+a3, зa+a3, 4a+a1, &c.

Ans. na+a”, and i̟n(n+1)a+aa®—1.

Find the nth term and the sum of n terms of each of the following series :

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(19.) A horse is sold for a farthing for the first nail in his shoes, 2 farthings for the next, 4 for the next, and so on, doubling for every nail; find the price of the horse. Ans. £4473924, 5s. 3 d.

(20.) Suppose it were possible to pay 1 grain of wheat for the first square of a chess-board, 2 for the next, 4 for the next and so on, doubling for each of the 64 squares; how many bushels would it amount to, allowing 7680 grains to the pint? And supposing Europe to contain 1,000,000 square miles of arable land, how many years would it take to grow it if the whole were in crop, and each acre produced on an average 40 bushels per annum?

Ans. 37529996894754 bush. and 1466 years.

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73

CHAPTER X.

PERMUTATIONS AND COMBINATIONS.

80. Permutations denote the different orders in which any quantities may be arranged; Combinations denote the different collections that may be formed out of them, without regard to the order in which they are placed. Thus ab and ba form different permutations, but only one combination.

be n.

81. If there be n things, a, b, c, d, &c., then, by taking them one by one, the number of permutations formed will By placing a before each of the other (n-1) things, there will be (n-1) permutations with a standing first; also in like manner (n-1) with b standing first; and so on for the whole n quantities. Hence there are

n(n-1) permutations taken two together.

Suppose a to be removed, there are (n-1) things remaining, and of these there are

(n − 1) (n − 2) permutations taken two together.

By prefixing a to each of these, there will be

(n-1) (n-2) permutations of three with a standing first; and similarly (n-1) (n-2) with b standing first; and so on for each of the n quantities. There are therefore

n(n-1) (n-2) permutations taken three together.

...

Collecting these results, and taking P1, P2, P ̧ Pr to denote the numbers of permutations of n things taken 1, 2, 3,... r, together respectively, we get

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-

Pn(n-1) (n − 2).

Similarly, it might be shown that

P1n(n-1) (n − 2) (n—3)

4

Pn(n-1) (n − 2) (n − 3) (n —4);

and hence we may infer, that generally

Prn(n-1) (n − 2) (n − 3). . (n − r + 1).

82. Assuming that this law holds true when (r — 1) things are taken together, then

Pr− 1 = n(n − 1) (n − 2)

but if one of the things, viz. a, be

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omitted, it must be equally true that the number of permutations of the remaining (n-1) things taken (r — 1) together, is

...

(n − 1) (n − 2) (n − 3) . . . (n − r + 1) found by putting (n − 1) for n in the preceding equation.

Now, by prefixing a to each of these last permutations, their number would remain the same, but there would be r taken together instead of r — 1.

In like manner it may be shown that the same number of permutations may be formed of these n things taken r together, in which b stands first, and so on for each of the others. If therefore

Pr−1 = n(n − 1) (n − 2) . . . . (n − r + 2),

it follows that

....

Pr = n(n − 1) (n − 2) (n − 3) .

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...

(n−r + 1).

That is, if the assumed law be true for any one value of r, it must be true for the next higher value. But it has been proved to hold true when r= 2, and r = 3; therefore it is true when r = 4, and if true for 4, then also for 5; and so on for any number.

If r = =n, the last factor in the above formula becomes 1, therefore denoting the number of permutations of n things taken altogether by Pn, we have

or

Pn = n(n-1) (n − 2)

=1.2.3.4

.

3.2.1,

n.

EXERCISES.

(1.) Find the number of permutations of 8 things taken 2 together, 3 together, 4 together, and all together.

Ans. 56, 336, 1680, and 40320. (2.) In how many different ways may a party of 12 be Ans. 479001600.

arranged at table?

1

(3.) The number of permutations of n things, taken 4 together = 8 times the number taken 3 together; find n.

1

Ans. 11.

(4.) In how many ways may a class of 9 boys be arranged? Ans. 362880. (5.) In the permutations formed out of the eight letters a, b, c, d, e, f, g, h, taken altogether, how many begin with ab? how many with abc? and how many with abcd?

Ans. 720, 120, and 24. (6.) The number of permutations of 18 things taken r together 12 times the number taken r 1 together; find r. Ans. 7. (7.) In how many ways can 8 persons be seated at a round table, so that all shall not have the same neighbours twice? Ans. 2520.

(8.) Find the number of changes which can be rung with a peal of 7 bells, and what difference will be made in the number by the absence of one ringer? Ans. 5040, & 4320.

83. Again, let C1, C2, C3 ... Cr, denote the number of combinations of n things taken 1 by 1, 2 by 2, 3 by 3 . . . r by r together;

then it is evident that C1 = n.

1

And since each combination of 2 things, as ab, admits of 2 permutations, ab, and ba, it follows that

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Similarly, since each combination of 3 things admits of 3.2.1 permutations,

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And, generally, since each combination of r things admits.

of 1.2.3... r permutations,

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