How many different 5-letter 'words' can be formed from the word 'statistics'?

I really am pretty stumped. I understand how to calculate more simpler questions in which each letter of the word is different using the Permutation formula n!/(n-r)! I can also deal with questions where there are repeating letters, but the 'new' words that are being created are the same length as the original word. But for this type of question where there are repeating letters and the 'new words' are shorter than the original, I don't know where to start. I don't need the precise answer - I just want to know how to get there. In fact, I know the answer (1390) but I cannot come up with a solution. I have tried using the permutation formula but it doesn't seem appropriate for this question.

  • $\begingroup$ What is your progress on this problem? What are your ideas? $\endgroup$ – Sasha Patotski Oct 3 '13 at 23:05

I will give a basic counting solution. If you want an answer using generating functions, look at the answer to this question.

If all five letters are different (which happens when you use one of each), you get $5!=120$ words.

If only two of the letters are equal (two 'S', two 'T', or two 'I', and $\binom{4}{3}$ ways to choose the remaining three), you have $\frac{5!}{2!}=60$ words for each of {'S', 'T', 'I'}. In total $720$ words.

If you have "two pairs" (there are three ways to get the pairs, and three ways to get the last letter in each case), there are $\frac{5!}{2!2!}=30$ words. In total for all these cases, there are $270$ words.

If you have "full house" (there are two ways to get three equal, and two ways for each of those to get the last pair), there are $\frac{5!}{3!2!}=10$ words. In total for these cases, there are $40$ words.

If you have three equal and two different (the triple can be had in two ways, and the remaining two can be had in $\binom{4}{2}=6$ ways for each triple), there are $\frac{5!}{3!}=20$ words. For all these cases, there are $240$ words.

Summing all of these cases gives $1390$ words.

  • $\begingroup$ The answer given to the question in the textbook was 1390. Your answer makes sense to me but it seems that some multiple is missing from the calculations... I can't see where though. $\endgroup$ – frantastic Oct 4 '13 at 0:41
  • $\begingroup$ @frantastic, I'm not in a position to review the argument carefully right now, but never discount the possibility that the textbook answer is wrong. Often, you will be able to find a list of errata for your textbook somewhere online (often on the writer's or publisher's website). $\endgroup$ – dfeuer Oct 4 '13 at 1:35
  • $\begingroup$ @frantastic: I hade made an error in the "pair" case, forgetting to count the number of ways to choose the last three. In the last case, I had gotten 120 instead of 240 through bad mental calculation. $\endgroup$ – Mårten W Oct 4 '13 at 8:29

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