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⭕️Algebra - Revision Notes on Permutations⭕️

➖The concept of permutation is used for the arrangement of objects in a specific order i.e. whenever the order is important, permutation is used.

➖The total number of permutations on a set of n objects is given by n! and is denoted as nPn = n!

➖The total number of permutations on a set of n objects taken r at a time is given by nPr = n!/ (n-r)!

➖The number of ways of arranging n objects of which r are the same is given by n!/ r!

➖If we wish to arrange a total of n objects, out of which ‘p’ are of one type, q of second type are alike, and r of a third kind are same, then such a computation is done as n!/p!q!r!

➖Al most all permutation questions involve putting things in order from a line where the order matters. For example ABC is a different permutation to ACB.

➖The number of permutations of n distinct objects when a particular object is not to be considered in the arrangement is given by n-1Pr

➖The number of permutations of n distinct objects when a specific object is to be always included in the arrangement is given by r.n-1Pr-1.

➖If we need to compute the number of permutations of n different objects, out of which r have to be selected and each object has the probability of occurring once, twice or thrice… up to r times in any arrangement is given by (n)r.

➖Circular permutation is used when some arrangement is to be made in the form of a ring or circle.

➖When ‘n’ different or unlike objects are to be arranged in a ring in such a way that the clockwise and anticlockwise arrangements are different, then the number of such arrangements is given by (n – 1)!

➖If n persons are to be seated around a round table in such a way that no person has similar neighbor then it is given as ½ (n – 1)!

➖The number of necklaces formed with n beads of different colors = ½ (n – 1)!

➖nP0 =1

➖nP1 = n

➖nPn = n!/(n-n)! = n! /0! = n! /1= n!

♦️Revision Notes on Arithmetic Progression♦️
If ‘a’ is the first term and ‘d’ is the common difference of the arithmetic progression, then its nth term is given by an = a+(n-1)d

The sum, Sn of the first ‘n’ terms of the A.P. is given by Sn = n/2 [2a + (n-1)d]

If Sn is the sum of n terms of an A.P. whose first term is ‘a’ and last term is ‘l’,Sn = (n/2)(a + l)

If common difference is d, number of terms n and the last term l, then Sn = (n/2)[2l-(n -1)d]

If a fixed number is added or subtracted from each term of an A.P., then the resulting sequence is also an A.P. and it has the same common difference as that of the original A.P.

If each term of A.P is multiplied by some constant or divided by a non-zero fixed constant, the resulting sequence is an A.P. again.

If a1, a2, a3, …, an andb1, b2, b3, …, bn, are in A.P. then a1+b1, a2+b2, a3+b3, ……, an+bn and a1–b1, a2–b2, a3–b3, ……, an–bn will also be in A.P.

Suppose a1, a2, a3, ……,an are in A.P. then an, an–1, ……, a3, a2, a1 will also be in A.P.

If nth term of a series is tn = An + B, then the series is in A.P.

If a1, a2, a3, ……, an are in A.P., then a1 + an = a2 + an–1 = a3 + an–2 = …… and so on.

In order to assume three terms in A.P. whose sum is given, they should be assumed as a-d, a, a+d.

Four terms of the A.P. whose sum is given should be assumed as a-3d, a-d, a+d, a+3d

Five convenient numbers in A.P. a–2b, a–b, a, a+b, a+2 b.

In general, we take a – rd, a – (r – 1)d, …., a – d, a, a + rd in case we have to take (2r + 1) terms in an A.P.

Likewise, any 2r terms of an A.P. should be assumed as: a – (2r-1)d, a – (2r – 3)d, …., a – d, a, a + d, ………….. , a+(2r-3)d, a + (2r-1)d.

The arithmetic mean of two numbers ‘a’ and ‘b’ is (a+b)/2.

The terms A1, A2, ….. , An are said to be arithmetic means between a and b if a, A1, A2, ….. , An, bis an A.P.

Clearly, ‘a’ is the first term, ‘b’ is the (n+2)th term and ‘d’ is the common difference. Then, we have b = a+(n+2-1)d = a+(n+1)d

Hence, this gives ‘d’ = (b-a)/(n+1)