a) In how many ways can $6$ people be lined up to get on a bus?

b) If $3$ specific persons, among $6$, insist on following each other, how many ways are possible?

My confusion

I understand that a) is just 6!, but apparently b) is $4! \cdot 3!$. Can anyone explain where they got 4! from? I thought it was $3! \cdot 3!$, since $3$ people are following each other, and $3$ aren't.

Thank you.

  • $\begingroup$ What a)? I don’t see any a) in your post. $\endgroup$
    – Aig
    Feb 24 at 18:11
  • $\begingroup$ What is the color of the bus? $\endgroup$
    – Dan Asimov
    Feb 24 at 19:18
  • $\begingroup$ Suppose it were only 4 people, and three of them insisted on getting on together. How many then? Is it $2!·2!$ as your answer suggests? If you're not sure of the correct answer, you can just write out all 24 possible orders, cross out the ones in which persons $A$ and $B$ don't get on together, and count how many orders are left. $\endgroup$
    – MJD
    Feb 25 at 3:11

1 Answer 1


The three people who insist on following each other can be considered as one unit (or person) and then along with the three remaining, they can be arranged in a line in $(3 + 1)! = 4!$. Doing this ensures that the $3$ people never get separated (always follow each other).

However, those three people can be arranged among themselves, so we need to multiply the result above by $3!$. And the final answer becomes $4! \times 3!$.

The problem with your approach $(3! \times 3!)$ is that it assumes the three special people must be placed in front (or behind) of the remaining three but does not account for the freedom to place them in either of the two positions. It also doesn't account for the fact that the group can be placed somewhere in between two of the remaining three people too.

To compensate for this, you can multiply by $4$ (there are four positions where the group can be placed) and you will end up with the correct result.

$$\times \; P_1 \times P_2 \times P_3 \; \times$$


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