Question

In how many ways can $6$ speakers A, B, C, D, E, F address a gathering if
$(i)$A speaks after B
$(ii)$A speaks immediately after B

Answer 1

First, we have to define what the terms we need to solve the problem are.
Since there are a total of six speakers to give a gathering of address and there are A, B, C, D, E, and F
And the question is to find the speaker A will need to speak only after B, also two restrictions are speaker A will need to speak after B in whatever the position like if B will speak at position one then A may speak position of two or three or something else, and speaker A will need to speak after B immediately means if the position of B is one means then A will need to be two.

Complete answer:
Since there are six speakers in the address gathering and we need to find $(i)$ A speaks after B
Let there is the total of five ways that the B can be placed first (on the position six B cannot be placed because after B, speaks A will need to speak and hence further) so there are $5$ways for the B can be placed in the gathering, so after placing B first, now there are only four ways out of four places for the speaker A hence $4!$ ways. (B placed at position one)
Similar for B placed at position two in four ways and A can be placed in $4!$ ways
B can be placed at position three in three ways and A can be placed is $4!$ ways, B can be placed in position four in two ways, and similarly, we get the sequence of $(5 \times 4 \times 3 \times 2 \times 1) \times 4!$ ways.
Hence A speaks after B is $360$ ways (using combination formula)

Or we can also able to apply this in a formula, number of ways of arranging; $(5 \times 4!) + (4 \times 4!) + (3 \times 4!) + (2 \times 4!) + (1 \times 4!)$ to get the same resultant
$(ii)$A speaks immediately after B
The only difference is A will need to speak immediately after B;
If B is at position one then A need to be position two $5 \times 4$
If B is at position two then A need to be position three $4 \times 3$ similar we can find for all position
Thus, we get A speaks immediately after B is $(5 \times 4) + (4 \times 3) + (3 \times 2) + (2 \times 1)$ ways
Hence there are $40$ ways.

Note: Since we used the method of combination is the number of ways to the given question,
Also, there is a concept like with repetition or without repetition is the restriction and similarly, here A will need to speak immediately after B will differ from A speaks after B. and hence the resultant will have some changes in the number of ways.