M1. Permutation and Combination:

When the order does matter, it is a Permutation. Just a collection without any regard to order or arrangement; it is a Combination.

1. Permutation:

The arrangement of objects in some order. 

a) Permutation of objects are all different: 

The total number of permutation of a set $n$, number of object taken $r$ at a time is given by:

$P(n,r)$ = $^{n}\textrm{P}_{r}$ = $\frac{n!}{(n-r)!}$ .......................................... (i)

If $r = n$; 

$P(n,n)$ = $n!$ .......................................... (ii) 

Where, $n! = 1.2.3.4 ................ (n-1).n$

Example: How many plates of vehicles consisting of $4$ different digits can be made out of the integers $4, 5, 6, 7, 8, 9$ . 

⇨ This is just like arranging $4$ objects out of $6$ integers. So we have 

     $P(6,4) = \frac{6!}{(6-4)!} = \frac{6.5.4.3.2.1}{2.1} = 360$

b) Permutation of objects are not all different: 

The total number of permutation of a set $n$ are taken, when $p$ of the objects are of first kind, $q$ of them are of second kind, $r$ of them are of the third kind, and remaining all are different. Then,

The total number of permutations = $\frac{n!}{p!\;q!\;r!}$ .................. (iii)

Example: In how many ways can the letters of the word 'CALCULUS' be arranged? 

⇨There are $8$ letters in the word 'CALCULUS'; But $C$ comes twice, $L$ comes twice, $U$ comes twice and remaining are different.
i.e. $n = 8$; $p = 2$; $q = 2$; $r = 2$ 

Number of ways in which the letters of the given word can be arranged $= \frac{n!}{p!\;q!\;r!}$ 

$= \frac{8!}{2!\;2!\;2!} = \frac{8*7*6*5*4*3*2*1}{2*1\;2*1\;2*1} = 5040$ 

c) Circular Permutation:
The total number of permutations of a set of $n$ objects arranged in a circle is $(n-1)!$

$P = \frac{n!}{n} = (n-1)!$ ................................ (iv)

Example: In how many ways can $7$ students be seated in a circle? 

⇨ Here, $n = 7$
The required number of ways = $(n-1)! = (7-1)! = 6! = 720$

2. Combination: 

The selection of objects without any regard to order (or, arrangement). 

The total number of combinations of $n$ objects taken $r$ at a time , Then the combination $C(n,r)$ is given as:

$^{n}\textrm{C}_{r}$ = $C(n,r)$ = $\frac{n!}{(n-r)!\;r!}$ .................................... (v)

Where, $n$ = total number of selection (combination)
             $r$ = different objects taken at a time

Example 1: A committee is to be chosen from $12$ men and $8$ women and is to consist of $3$ men and $2$ women. How many committees can be formed?

⇨ The number of ways choosing $3$ men from $12$ men is $C(12, 3) = 220$
    The number of ways choosing $2$ women from $8$ women is $C(8,2) = 28$ 

∴ Total nos of Committees formed $= 220 * 28 = 6160$

Example 2: A committee of $5$ persons is to be selected from $5$ men and $4$ ladies. In how many ways can this be done so that at least $2$ ladies are always included. 

⇨ The committee may consist of $3$ men and $2$ ladies, or $2$ men and $3$ ladies, or $1$ men and $4$ ladies. Then from these condition: 

The number of ways forming the committee consisting of $3$ men and $2$ ladies = $C(5,3) * C(4,2)$ 

= $\frac{5!}{(5-3)!\;3!} * \frac{4!}{(4-2)!\;2!} = \frac{5!}{2!\;3!} * \frac{4!}{2!\;2!} = \frac{5*4}{2} * \frac{4*3}{2} = 10 * 6 = 60\;ways$

The number of ways forming the committee consisting of $2$ men and $3$ ladies = $C(5,2) * C(4,3)$ 

= $\frac{5!}{(5-2)!\;2!} * \frac{4!}{(4-3)!\;3!} = \frac{5!}{3!\;2!} * \frac{4!}{1!\;3!} = \frac{5*4}{2} * \frac{4}{1} = 10 * 4 = 40\;ways$

The number of ways forming the committee consisting of $1$ men and $4$ ladies = $C(5,1) * C(4,4)$ 

= $\frac{5!}{(5-1)!\;1!} * \frac{4!}{(4-4)!\;4!} = \frac{5!}{4!\;1!} * \frac{4!}{0!\;4!} = \frac{5}{1} * \frac{1}{1} = 5\;ways$

∴ Total number of ways $= 60 + 40 + 5 = 105\;ways$

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