Permutations and Combinations

Correct answer is option (a) 24 There are 5 speakers - A, B, C, D & E. A will speak always before B. So, we will consider these two speakers A and B as one group. And consider other 3 speakers as individual groups. So, there are 4 groups that have to be arranged in a row. So, it can be done in 4P4 different ways. Hence, The no. of ways in which A will speak before B = 4P4 = 4 x 3 x 2 x 1 = 24

Varshaa M

CA Final

★ 35K+

03-Jun-23 16:01

Thank You sister..😀

Thread Starter

Sr the Unique

CA Inter

★ 10

03-Jun-23 16:21