Question

if 5 men a, b, c, d, e intend to speak at a meeting, in how man orders can they do so without b speaking before a? how many orders are there in which a speaks immediately before b?

Answer 1

We calculate the number of ways 5 men can speak at a meeting without B speaking before A by treating A and B as a single entity and then arranging, resulting in 4! permutations.

Step-by-step explanation:

The question deals with permutations, which is a concept in combinatorics, a branch of mathematics concerning counting. To find the number of orders in which 5 men can speak at a meeting without person B speaking before person A, we need to consider that once A has spoken, the remaining people can speak in any order.

There are 5! (factorial) total ways to arrange 5 people, but since we want orderings with A before B, we consider the pair A and B as a single entity and then arrange.

So, we have 4! ways to arrange the four entities (A and B considered as one, and C, D, E) and then 2! ways to arrange A and B within that single entity - since A has to come before B, it's actually only 1 way. Multiplying 4! by 1 gives us the total number of ways without B before A.

To calculate the number of orders in which a speaks immediately before b, we need to fix the positions of a and b as ab. Then the other 3 men can be arranged in 3! ways. So, the number of orders in which a speaks immediately before b is 3!.