Question

A meeting is to be addressed by 5 speakers A, B, C, D and E. In how many ways can the speakers be ordered, if B must not precede A (immediately or otherwise)?
A. 120
B. 24
C. 60
D. ${5^4} \times 4$

Answer 1

Hint: Here, we will find the possible number of ways by putting A at different places with the given condition in question and add all the answers of different cases to get the total number of possible ways.

Complete step by step answer
Given, total number of speakers are 5 i.e. A, B, C, D and E.
Case I:
A speaks first, this is possible in 4! ways = $4 \times 3 \times 2 \times 1 = 24$ ways
Case II:
A speaks second, then at first 3 ways are possible i.e. (C, D and E) as we are given B must not precede A (immediately or otherwise) and after A 3! ways are possible.
Possible number of ways in this case = $3 \times 3! = 3 \times 6 = 18$ ways
Case III:
A speaks third, then the possible number of ways at first two places is $\left( {3 \times 2} \right)$ ways = 6 ways and possible ways at two places after A is 2!.
Possible ways in this case = $6 \times 2 = 12$ ways
Case IV:
A speaks fourth, then possible number ways at first three places are $\left( {3 \times 2 \times 1} \right)$ ways and possible ways at fifth place is 1.
Possible ways in this case = 6
And A cannot speak at last because B must not precede A (immediately or otherwise).
Total number of possible ways = $24 + 18 + 12 + 6 = 60$ ways
Option C is correct among all the given options.

Note: This question is related to permutations, therefore, in these types of questions; find all the possible cases independently, considering the given situation in question. Always remember ‘and’ in permutation means multiplication and ‘or’ in permutation means addition.