# Permutations & Combinations - Lesson 2

Hi friends, in my previous introductory lesson of Permutations & Combinations, we have learnt some fundamentals of Permutations & Combinations. Today we shall learn Lesson 2 of Permutations & Combinations. Before reading this, please read the Lesson 1 from HERE.

### Circular Permutations

When we consider the arrangements of objects in a line, such permutations are known as linear permutations instead of arranging the objects in a line, if we arrange tem in the form of a circle, we can call them circular permutations.

In circular permutations, objects are arranged along the circumference of a circle. Here, there is neither a beginning nor an end. We fix the position of one objects and then arranger the remaining (n-1) objects in all possible ways.

This can be done in (n-1)!

Note :
1. The number of circular permutations of n different objects = (n-1)!
2. The number of ways in which n person can be seated around a circular table is (n-1)!

### SOLVED EXAMPLES

1. In how many can 10 businessmen sit around a round table such that 2 businessmen
(i) Always sit together ii) never sit together

Sol.

(i) If 2 businessmen always sit together we can consider them to be one unit. Thus, we have to arrange 9 units in a circle. This can be done in (9 — 1)! = 8! ways.

Now, the two businessmen can sit in 2! Ways between themselves. Hence, total number of way

= 2 x 8!

(ii) Number of ways in which two businessmen never sit together

=9!-2x8!(9-2) = 7x8!

2. In how many ways can 5 boys and 5 girls be made to sit around a table with 10 chairs so that no two boys and no two girls sit side by side ?

Sol.

Given there are 10 chairs boys 5! Ways (BBBB2) so that five girls can be arranged in 5! Ways in the five gaps created by the boys.

On the other hand, the sequence or the line can start with a boy (or) a girl first.

Therefore, total arrangement = 2x 5!x 5!.

### Combinations or Selections

If n objects are given and we have to choose r(r
nCr = n!  / r!(n-r)!

### Difference between Permutation and Combination

Note

1. nC0 = 1 and nCn = 1
2. nCp = nCq, p+q = n or p=q
3. nCr = nCn-r
4. nCr-1 + nCr = n+1 Cr
5. Number of combination of 'n' different things, taken 'g' at a time when, 'p' particular things always occur n-p C r-p
6. Number of combinations of 'n' different things taken 'r' at a time when, 'p' particular things never occur = n-p C r

#### SOLVED EXAMPLES

1. Four cards are to be selected from a pack. In how many ways can this be done so that the selection has all hearts ?

Sol

In a pack of cards there are 13 each of spades, clubs, diamonds and hearts. 4 hearts can be chosen from 13 in 13C4

2. There are 15 boys and 10 girls. A committee of 3 boys and 2 girls is to be formed. Provided that a particular boy must be included, fiend the number of ways the selection can be done ?

Sol. Since one boy must be included, the other 2 can be chosen out of 14 in 14C2 ways = 9!

The girls can be chosen out of 14 in 14C2 ways = 9!

The girls can be chosen in 10C2 ways = 45

Hence, the number of ways = 91 x 45 = 4095.

3. In how many ways can the letters of the word 'BETTER' be arranged such that the T's are always together ?

Sol

For the two T's to be together we can consider them to be 1 unit instead of two. Hence, there are 5! Arrangements. Also there are two E's

Hence the answer is 5! / 2! = 60 .

4. When two cards are drawn at random from a pack of 52 cards, in how many of the outcomes will the sum of two cards add up to be odd number ?

Sol

The total number of numbered cards in a pack is equal to 36.

=> No. of ways of selecting two cards in a pack from 36 cards = 36C2

When 2 is added to 3. . . . . . .9 individually, we get 4 odd numbers and 3 even numbers.

Similarly, we get the same result for 3,4,......... ,9.

=> probability of getting the sum of the cards as an odd number is 4 / (4+3) = 4/7

=> Total number of outcomes = (4/7) x 36C2

5. Arun has been given two baskets, one of which is empty and the other is filled with 20 balls of identical size. Out of the 20 balls, one is red coloured, one is green coloured, one is black coloured and the remaining are white coloured and the remaining are white coloured. He is asked to put all the balls into the empty basket one after another such that red ball should be put before the green one and green should be put before the black ball. What is the total number of ways in which Arjun can do the work ?

1. 20!/6
2. 20!/6!
3. 20!/7!
4. None of these.
Sol. Arun can select positions for red, black and green balls in 20C3 ways and the remaining 17 balls can be selected in 17! Ways.

Hence, the total number of ways is 20C3 x 17! = 20!/ 3! = 20! / 6

shared by Aindree Mukherjee