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Find the series in which 1
1 5. The 15th term is and the 23rd term is 25
4 6. The 2nd term is 2, and the 31st term is
1 7. The 39th term is and the 54th term is 11
26 Find the harmonic mean between
1 1 8. 2 and 4. 9. 1 and 13.
4 10 1 1
1 11. and 12. and
13. x+y and x- y. a ū
x+y 1-Y 14. Insert two harmonic means between 4 and 12. 15. Insert three harmonic means between 27 and 12. 16. Insert four harmonic means between 1 and 6. 17. If G be the geometric mean between two quantities A and B,
show that the ratio of the arithmetic and harmonic means of A and G is equal to the ratio of the arithmetic and
harmonic means of G and B. 18. To each of three consecutive terms of a G.P. the second of
the three is added. Show that the three resulting quantities
are in H.P. Sum the following series : 19. 1+13+3+...... to 6 terms, , 20. 1+14+2} + ...... to 6 terms. 21. (2a+x)+30+(4a-x)+ ...... to p terms.
4 22. 16 - 1}+
to 8 terms. 5
to 12 terms.
the harmonic mean between y - x and y– 2. If a, b, c, d be in A.P., a, e, f, d in G.P., a, g, h, d in H.P.
respectively; prove that ad=ef=bh=cg. If a?, 62, c2 be in A.P., prove that b+c,c+a, a+b are in H.P.
PERMUTATIONS AND COMBINATIONS.
365. Each of the arrangements which can be made by taking some or all of a number of things is called a permutation.
Each of the groups or selections which can be made by taking some or all of a number of things is called a combination.
Thus the permutations which can be made by taking the letters a, b, c, d two at a time are twelve in number: namely,
ab, ac, ad, bc, bd, cd,
ba, ca, da, cb, db, dc; each of these presenting a different arrangement of two letters.
The combinations which can be made by taking the letters a, b, c, d two at a time are six in number: namely,
ab, ac, ad, bc, bd, cd; each of these presenting a different selection of two letters.
From this it appears that in forming combinations we are only concerned with the number of things each selection contains; whereas in forming permutations we have also to consider the order of the things which make up each arrangement; for instance, if from four letters a, b, c, d we make a selection of three, such as abc, this single combination admits of being arranged in the following ways:
abc, acb, bca, bac, cab, cba, and so gives rise to six different permutations.
366. Before discussing the general propositions of this chapter the following important principle should be carefully noticed.
If one operation can be performed in m ways, and (when it has been performed in any one of these ways) a second operation can then be performed in n ways; the number of ways of performing the two operations will be mxn.
If the first operation be performed in any one way, we can associate with this any of the n ways of performing the second operation: and thus we shall have n ways of performing the two operations without considering more than one way of performing the first; and so, corresponding to each of the m ways of performing the first operation, we shall have n ways of performing the two; hence altogether the number of ways in which the two operations can be performed is represented by the product
m X n.
Example. Suppose there are 10 steamers plying between New York and Liverpool; in how many ways can a man go from New York to Liverpool and return by a different steamer ?
There are ten ways of making the first passage; and with each of these there is a choice of nine ways of returning (since the man is not to come back by the same steamer); hence the number of ways of making the two journeys is 10 x 9, or 90.
This principle may easily be extended to the case in which there are more than two operations each of which can be performed in a given number of ways.
367. To find the number of permutations of n dissimilar things taken r at a time.
This is the same thing as finding the number of ways in which we can fill up r places when we have n different things at our disposal
The first place may be filled up in n ways, for any one of the n things may be taken; when it has been filled
any one of these ways, the second place can then be filled up in 1 – 1 ways; and since each way of filling up the first place can be associated with each way of filling up the second, the number of ways in which the first two places can be filled up is given by the product n(n-1). And when the first two places have been filled up
in any way, the third place can be filled up in n-2 ways. And reasoning as before, the number of ways in which three places can be filled up is n (n − 1) (n − 2).
Proceeding thus, and noticing that a new factor is introduced with each new place filled up, and that at any stage the number of factors is the same as the number of places filled up, we shall have the number of ways in which r places can be filled up equal to
n(n-1) (n − 2)......to r factors; and the goth factor is n - (1 - 1), or n-r+l.
Therefore the number of permutations of n things taken r at a time is
n(n-1)(n-2)......(n-r+1). CoR. The number of permutations of n things taken all at a time is
n(n-1)(n--2)......to n factors,
n (n − 1) (n— 2)... It is usual to denote this product by the symbol in, which is read “factorial n.” Also n! is sometimes used for In.
368. We shall in future denote the number of permutations of n things taken r at a time by the symbol "P,, so that
=n (n − 1) (n − 2)......(n-r+1); also
In working numerical examples it is useful to notice that the suffix in the symbol "P, always denotes the number of factors in the formula we are using.
Example 1. Four persons enter a carriage in which there are six seats ; in how many ways can they take their places ?
The first person may seat himself in 6 ways; and then the second person in 5; the third in 4; and the fourth in 3; and since each of these ways may be associated with each of the others, the required answer is 6 x 5 x4x3, or 360.
Example 2. How many different numbers can be formed by using six out of the nine digits 1, 2, 3,...9?
Here we have 9 different things and we have to find the number of permutations of them taken 6 at a time; ... the required result='P6
= 60480. 369. To find the number of combinations of n dissimilar things taken r at a time.
Let nC, denote the required number of combinations.
Then each of these combinations consists of a group of y dissimilar things which can be arranged among themselves in ways. [Art. 367, Cor.]
Hence "Cox /r is equal to the number of arrungements of n things taken r at a time; that is, "Cor="P,=n(n-1)(n-2)...(n-+1);
n(n-1)(n-2)...(n-r+1) .; "Cr=
Cor. This formula for "C, may also be written in a different form; for if we multiply the numerator and the denominator by
r we obtain
n (12-1)(n-2)... (n-r+1) * In -
Irin since n (n-1) (1 – 2) (n-r+1) n – r= n.
Example. From 12 books in how many ways can a selection of 5 be made, (1) when one specified book is always included, (2) when one specified book is always excluded ?
(1) Since the specified book is to be included in every selection, we have only to choose 4 out of the remaining 11.
11 x 10 x 9x8 Hence the number of ways=11C4
1x2x3 x4 (2) Since the specified book is always to be excluded, we have to select the 5 books out of the remaining 11.
11 x 10 x 9 x 8 x 7 Hence the number of ways=11Cg=
1x2x3 x4 x5 370. The number of combinations of n things r at a time is equal to the number of combinations of n things n -r at a time.
In making all the possible combinations of n things, to each group of r things we select, there is left a corresponding group of n-y things; that is, the number of combinations of n things r at a time is the same as the number of combinations of n things n-r at a time;
.. "Cr="CRThis result is frequently useful in enabling us to abridge arithmetical work.
Example. Out 14 men in how many ways can an eleven be chosen ? The required number=14Cu
14 x 13 x 12
1x2x3 If we had made use of the formula 14Cu, we should have had to reduce an expression whose numerator and denominator each contained 11 factors.
371. In the examples which follow it is important to notice that the formula for permutations should not be used until the suitable selections required by the question have been made.