# PERMUTATION & COMBINATION

**PERMUTATION & COMBINATION**

**FUNDAMENTAL PRINCIPLE:**

If an event A can occur in 'm' different ways and another event B can occur in 'n' different ways, then the total number of different ways of-

(a) simultaneous occurrence of both events in a definite order is m n. This can be extended to any number of events (known as multiplication principle).

(b) happening exactly one of the events is m + n (known as addition principle).

**Example :**

A college offers 6 courses in the morning and 4 in the evening. The possible number of choices with the student if he wants to study one course in the morning and one in the evening is?

**Solution :**

The student has 6 choices from the morning courses out of which he can select one course in 6 ways.

For the evening course, he has 4 choices out of which he can select one in 4 ways.

Hence the total number of ways 6 * 4 = 24.

**Example :**

A college offers 6 courses in the morning and 4 in the evening. The number of ways a student can select exactly one course, either in the morning or in the evening?

**Solution :**

The student has 6 choices from the morning courses out of which he can select one course in 6 ways.

For the evening course, he has 4 choices out of which he can select one in 4 ways.

Hence the total number of ways 6 + 4 = 10.

**PERMUTATION & COMBINATION :**

**Factorial :**

A Useful Notation : n! = n.(n – 1).(n – 2)..............3. 2. 1;

n! = n. (n – 1)! where n is real number.

**Points to Remember :**

- 0! = 1! = 1
- Factorials of negative integers are not defined.

- (2n)! = 2^n.n! [1. 3. 5. 7........(2n – 1)]

**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.

denotes the number of permutations of n different things, taken r at a time (n and r real numbers and r<= n)

**Points to Remember:**

Number of arrangements of n distinct things taken all at a time = n!

**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 group is called a COMBINATION.

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

nCr denotes the number of combinations of n different things taken r at a time (n => N, r => W, r < n)

**Points to remember :**

**Example :**

If a denotes the number of permutations of (x + 2) things taken all at a time, b the number of permutations of x things taken 11 at a time and c the number of permutations of (x – 11) things taken all at a time such that a = 182 bc, then the value of x is?

**Solution :**

**Example :**

A box contains 5 different red and 6 different white balls. In how many ways can 6 balls be drawn so that there are atleast two balls of each colour ?

**Solution :**

The selections of 6 balls, consisting of atleast two balls of each colour from 5 red and 6 white balls, can be made in the following ways

Therefore total number of ways = 425

**Example :**

If all the letters of the word 'RAPID' are arranged in all possible manner as they are in a dictionary,

then find the rank of the word 'RAPID'.

**Solution :**

First of all, arrange all letters of given word alphabetically : 'ADIPR'

No of word starting with A_ _ _ _ = 4! = 24

No of word starting with D_ _ _ _ = 4! = 24

No of word starting with I_ _ _ _ = 4! = 24

No of word starting with P_ _ _ _ = 4! = 24

No of word starting with RAD_ _ = 2! = 2

No of word starting with RAI_ _ = 2! = 2

No of word starting with RAID _ = 1! = 1

No of word starting with RAPI _ = 1! = 1

Rank of WORD 'RAPID' = 24+ 24 + 24 +24 + 2 + 2 + 1 + 1 = 102

**PERMUTATIONS OF ALIKE OBJECTS**

The number of permutations of n things taken all at a time : when p of them are similar of one type, q of them are similar of second type, r of them are similar of third type and the remaining n – (p + q+ r) are all different is

**Example :**

In how many ways the letters of the word "ARRANGE" can be arranged without altering the relative position of vowels & consonants.

**Solution :**

The consonants in their positions can be arranged in 4!/2! = 12 ways.

The vowels in their positions can be arranged in 3!/2! = 3 ways

so Total number of arrangements = 12*3 = 36

**Example :**

How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits always occupy the odd places?

**Solution :**

There are 4 odd digits (1, 1, 3, 3) and 4 odd places (first, third, fifth and seventh). At these places

the odd digits can be arranged in 4!/2!*2! = 6 ways

Then at the remaining 3 places, the remaining three digits (2, 2, 4) can be arranged in 3!/2! = 3 ways

The required number of numbers = 6*3 = 18

**Example :**

Find the total number of 4 letter words formed using four letters from the word ''PARALLELOPIPED'.

**Solution :**

Given letters are PPP, LLL, AA, EE, R, O, I, D.