Permutation and Combination Part II

Permutation and Combination Examples – UGC NET UNIT 1

Question 1: In how many ways 6 children can be arranged in a line such that

(i) Two particular children of them are always together

(ii) Two particular children of them are never together

Solution:

(i) 2 students need to be together, hence we can consider them 1.

Thus the remaining 5 gives the arrangement of 5! ways.

ie 5! = 5*4*3*2*1

= 120 ——————–(1)

also, two children in a line can be arranged in 2! ways ————-(2)

Hence, the total number of arrangements = 120*2 = 240 ways

(ii) Total number of arrangements of 6 children will be 6!

ie 6! = 6*5*4*3*2*1

= 720 ways ———————-(1)

Tow children together can be arranged in 240 ways ———(2) (Ref (i))

Therefore, Two particular children are never together will be = 720 – 240 = 480 ways

Question 2: It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?

Solution

5 men and 4 women ie. Total 9 positions

Four places can be occupied by 4 women in P(4,4) ways = 4!

= 4*3*2*1

= 24 ways

Remaining 5 positions can be occupied by 5 men ie 5!

= 5*4*3*2*1

= 120 ways

Therefore, number of ways of seating arrangements = 24*120

= 2880 ways

Example 3: Find the number of words with or without meaning, that can be formed with the letters of the word SWIMMING?

Solution

SWIMMING = 8 Letters

here, I comes 2 times

& M comes 2 times

Therefore number of words formed = 8! /(2!*2!)

= 8*7*6*5*4*3*2*1/((2*1)*(2*1))

= 8*7*6*5*2*3

= 10080