Question

give answer in 4 decimal places In how many ways can 6 people be lined up to get on a bus if 2 specific persons, among 6, refuse to follow each other?

Answer 1

Explanation:

normally, for the possibilities of 6 people lining up we would have the permutations of 6 elements.

for the first we have 6 choices, for the second then only 5, for the third 4, for the fourth 3, for the fifth 2 and for the sixth then only one remaining choice.

that is

6! = 6×5×4×3×2×1 = 720

now in our special case, 2 people do not want to stand next to each other in the line.

there are many ways to model this.

my favorite would be to first take one of these two enemies out, and we line up the 5 remaining people without any problems.

that is 5! = 5×4×3×2×1 = 120

and now, for each of these 120 possibilities I mix in the 6th waiting person. I can't put him in front or behind his enemy. that leaves me with 4 choices for each of the 120 possibilities.

so, we have exactly

120×4 = 480 possible ways to line them up.

4 decimal places : 480.0 or 4.800 × 10²