I'm studying and reading through my discrete math book.. I seemed to be grasping the idea of permutations, but I don't understand how the solution for this particular problem came to be.


How many permutations of the letters ABCDEFGH contain the string ABC ?

Solution: Because the letters ABC must occur as a block, we can find the answer by finding the number of permutations of six objects, namely, the block ABC and the individual letters D, E, F , G, and H . Because these six objects can occur in any order, there are 6! = 720 permutations of the letters ABCDEFGH in which ABC occurs as a block.

My reasoning?

I know the permutation formula is: $\frac{n!}{(n-r)!}$ and there are 8 letters. We want ABC, so r=3? Using this formula.. wouldn't you get: $\frac{8!}{(8-3)!}$ = $\frac{8!}{5!}$ = 336?

I guess since ABC is a "block" then ABC is kind of like a single letter? so 8-3=5.. then +1 for the [abc] combo? Like I think I get that after looking at the solution, but If the solution wasn't provided I don't understand how you would know to approach it that way.

I'm probably just lacking common sense and this is a dumb question, but is this just something that should be common sense?



$n P k = \frac{n!}{(n-k)!}$ tells you how many ways you can line up $k$ out of $n$ distinct objects with the order mattering, so $n P k$ does not make sense here. Since you need ABC to appear in order, you can treat it as one block. Otherwise, you have to consider all the orderings of ABC... and then figure out which ones contain ABC as a block and normalize by that.

Alternatively, treat ABC as one "super letter" and then you just consider orders of {ABC,D,E,F,...} which is easily seen to be 6!.

  • $\begingroup$ Ok so using the formula is irrelevant for this question? I get the supper letter thing, but it can't be expressed in terms of n and k? $\endgroup$ – Cozen Mar 22 '14 at 22:45

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