Permutation and Combination Formula – Properties

Here you will learn formula permutation and combination and properties of permutation and combination with examples.

Let’s begin –

Permutation and Combination Formula

Permutation

Each of the arrangements in a definite order which can be made by taking some or all of the things at a time is called a PERMUTATION. In permutation, order of appearance of things is taken into account; when the order is changed, a different permutation is obtained.

Generally, it involves the problems of arrangements (standing in a line, seated in a row), problems on digit, problems on letters from a word etc.

\(^{n}P_r\) denotes, the number of permutations of n different things, taken r at a time (n \(\in\) N, r \(\in\) W, r \(\le\) n)

\(^{n}P_r\) = n(n – 1)(n – 2)……..(n – r + 1) = \(n!\over (n-r)!\)

Note :

(i) \(^{n}P_n\) = n!, \(^{n}P_0\) = 1, \(^{n}P_1\) = n

(ii) Number of arrangements of n distinct things taken all at a time = n!.

(iii) \(^{n}P_r\) is also denoted by P(n,r).

Combination

Each of the groups or selections which can be made by taking some or all of the things without considering the order of the things in each groups is called a COMBINATION.

Generally, involves the problem of selections, choosing, distributed groups formation, committee formation, geometrical problems etc.

\(^{n}C_r\) denotes the number of combinations of n different things taken r at a time (n \(\in\) N, r \(\in\) W, r \(\le\) n)

\(^{n}C_r\) = \(n!\over r!(n-r)!\)

Note :

(i)  \(^{n}C_r\) is also denoted by \(\binom{n}{r}\) or C(n,r).

(ii)  (ii) \(^{n}P_r\) = \(^{n}C_r\).r!

Example : How many four letter words can be formed from the letters of the word ‘ANSWER’ ? How many of these words start with vowel ?

Solution : Number of ways of arranging 4 different letters from 6 different letters are

      \(^{6}C_4\).4! = 360

      There are two vowels (A & E) in the word ‘ANSWER’

      Total number of 4 letter words starting with A : A _ _ _ = \(^{5}C_3\).3! = 60

      Total number of 4 letter words starting with E : E _ _ _ = \(^{5}C_3\).3! = 60

      \(\therefore\)    Total number of 4 letter words starting with a vowel = 60 + 60 = 120

Properties of Permutation and Combination

(a)  The number of permutation of n different objects taken r at a time, when p particular objects are always to be included is r!.\(^{n-p}C_{r-p}\) (\(p\le r\le n\)).

(b)  The number of permutation of n differnt objects taken r at a time, when repetition is allowed any number of times is \(n^r\).

(c)  Following properties of \(^{n}C_r\) should be remembered:

(i)  \(^{n}C_r\) = \(^{n}C_{n-r}\); \(^{n}C_0\) = \(^{n}C_n\) = 1

(ii)  \(^{n}C_x\) = \(^{n}C_y\) \(\implies \) x = y or x + y = n

(iii)  \(^{n}C_r\) + \(^{n}C_{r-1}\) = \(^{n+1}C_r\) 

(iv)  \(^{n}C_0\) + \(^{n}C_1\) + \(^{n}C_2\) + ………. + \(^{n}C_n\) = \(2^n\)

(v)  \(^{n}C_r\) = \(n\over r\)\(^{n-1}C_{r-1}\)

(vi)  \(^{n}C_r\) is maximum when r = \(n\over 2\) if n is even & r = \(n-1\over 2\) or r = \(n+1\over 2\), if n is odd.

(d)  The numbers of combinations of n different things taken r at a time,

(i)  when p particular things are always to be included = \(^{n-p}C_{r-p}\)

(ii)  when p particular things are always to be excluded = \(^{n-p}C_r\)

(iii)  when p particular thing are always to be included and q particular things are to be excluded = \(^{n-p-q}C_{r-p}\)

Example : There are 6 pockets in the coat of a person. In how many ways can he put 4 pens in these pockets?

Solution : First pen can be put in 6 ways.

Similarly each of second, third and fourth pen can be put in 6 ways

Hence total number of ways = \(6\times 6\times 6\times 6\) = 1296.

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