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**Permutation and combination Important formulas:**

**1. Factorial Notation:**

Let n be a positive integer. Then, factorial n, denoted n! is defined as:

n! = n(n – 1)(n – 2) … 3.2.1.

**Examples:**

We define 0! = 1.

4! = (4 x 3 x 2 x 1) = 24.

5! = (5 x 4 x 3 x 2 x 1) = 120.

**2. Permutations:**

The different arrangements of a given number of things by taking some or all at a time, are called permutations.

**Examples:**

**i.** All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).

**ii.** All permutations made with the letters a, b, c taking all at a time are:

( abc, acb, bac, bca, cab, cba)

**3.Number of Permutations:**

Number of all permutations of n things, taken r at a time, is given by:

n_{P}_{r} = n(n – 1)(n – 2) … (n – r + 1) = n! / (n – r)!

**Examples:**

**i.** 6_{P2} = (6 x 5) = 30.

**ii.** 7_{P3} = (7 x 6 x 5) = 210.

**iii.** Cor. number of all permutations of n things, taken all at a time = n!.

**4. An Important Result:**

If there are n subjects of which p_{1} are alike of one kind; p_{2} are alike of another kind; p_{3} are alike of third kind and so on and p_{r} are alike of r^{th} kind,

such that (p_{1} + p_{2}+ … p_{r}) = n.

Then, number of permutations of these n objects is = n! /(p_{1}!).(p_{2})!…..(p_{r}!)

**5.Combinations:**

Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.

**Examples:**

**i.** Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.

**Note:** AB and BA represent the same selection.

**ii.** All the combinations formed by a, b, c taking ab, bc, ca.

**iii.** The only combination that can be formed of three letters a, b, c taken all at a time is abc.

**iv.** Various groups of 2 out of four persons A, B, C, D are:

AB, AC, AD, BC, BD, CD.

Note that ab ba are two different permutations but they represent the same combination.

**6.Number of Combinations:**

The number of all combinations of n things, taken r at a time is:

n_{Cr} =n!/(r!)(n – r)! = n(n – 1)(n – 2) … to r factors / r!.

**Note:**

i. n_{Cn} = 1 and n_{C}_{0} = 1.

ii. n_{Cr} = n_{C}_{(n – r)}

Examples:

i. 11_{C4} = (11 x 10 x 9 x 8) /(4 x 3 x 2 x 1) = 330.

ii. 16_{C13} = 16_{C(16 – 13)} = 16_{C3}

= (16 x 15 x 14) /3! = (16 x 15 x 14) /(3 x 2 x 1) = 560

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