Permutation and Combination - Important Formulas

Factorial Notation:

Let n be a positive integer. Then, factorialn, 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.

Permutations:

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

Examples:

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

All permutations made with the letters a, b, c taking all at a time are:
( abc, acb, bac, bca, cab, cba)

Number of Permutations:

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

ⁿP_r = n(n - 1)(n - 2) ... (n - r + 1) =n!(n - r)!

Examples:

⁶P₂ = (6 x 5) = 30.

⁷P₃ = (7 x 6 x 5) = 210.

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

An Important Result:

If there are n subjects of which p₁ are alike of one kind; p₂ are alike of another kind; p₃ are alike of third kind and so on and p_r are alike of r^th kind, 
such that (p₁ + p₂ + ... p_r) = n.

Then, number of permutations of these n objects is =n!(p₁!).(p₂)!.....(p_r!)

Combinations:

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

Examples:

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.

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

The only combination that can be formed of three letters a, b, ctaken all at a time is abc.

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.

Number of Combinations:

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

ⁿC_r =n!=n(n - 1)(n - 2) ... to r factors.(r!)(n -r)!r!

Note:

ⁿC_n = 1 and ⁿC₀ = 1.

ⁿC_r = ⁿC_{(n - r)}

Examples:

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

ii.   ¹⁶C₁₃ = ¹⁶C_{(16 - 13)} = ¹⁶C₃ =16 x 15 x 14=16 x 15 x 14= 560.3!3 x 2 x 1