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So I've recently been reviewing problems and formulas for the upcoming AMC8 which I will be taking. Most of the stuff is pretty simple, but I've forgotten what formula I am able to use for this type of problem. Now I know how to deal with a few similar problems on it, but I'm not sure how to relate them back to this. The problem goes a little like this:

Suppose you have the world "BUBBLES" where you are allowed to mix the letters as you want. Each different permutation of letters counts as a distinct word if it cannot be recognized as a previous word already counted. Words do not necessarily have to match up with standard English words either, so "words" like "SULBEBB" or "BSLBEBU" are acceptable. How many four-letter permutations can be formed from the word "BUBBLES"?

Now I already know how to deal with the problem if all letters in the world BUBBLES were unique. If we had something like ABCDEFG, all that needs to be done is $\frac{7!}{3!} = 7*6*5*4 = 840$. But the issue comes in with the $3$ letters that are the same, which in this case is the 3 Bs. Now if we had to include the $3$ Bs no matter what, I know that we can simply divide again by $6!$ as that's the amount of ways that we would have overcounted the Bs for each scenario. But the issue comes in with the fact that not all the Bs need to be included. I'm not sure how to get pass this condition, does anyone have any ideas on it? And does anyone know a formula (and the way to derive it would be nice to hear as well, if you have one) or method I could use for this type of problem (excluding brute forcing it, of course)? Or do I have to casework through it manually?

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Let $T(n)$ denote the # of words possible, if $n$ B's are used : $n \in \{0,1,2,3\}.$

Then the total # words is $\sum_{n=1}^3 T(n)$.

$T(0) = 4!$.

$T(1) = \binom{4}{3} \times 4!$.

$T(2) = \binom{4}{2} \times 4 \times 3.$

$T(3) = \binom{4}{3} \times 4.$

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  • $\begingroup$ Yeah I figured that would be a way to do it. I did a pretty similar method of it, and got 136 in total as well. But I figured that this would kind of be under "casework" as well. $\endgroup$ – Boris Poris Oct 25 '20 at 3:11
  • $\begingroup$ @BorisPoris If the numbers had been larger (i.e. 17 B's and one letter each of the other 25 in the alphabet, and each word has 12 letters, for example) then I would have pursued elegant formulas. Since the numbers were small, there was no need to. $\endgroup$ – user2661923 Oct 25 '20 at 3:17
  • $\begingroup$ Oh then would you mind showing a formula for an example like that? I know that this problem had a particularly small sample, but I was looking for a more generalized formula for a larger word like you mentioned. So if we did end up having the example you gave (i.e. 17 B's and one letter each of the other 25 in the alphabet, and each word has 12 letters), then how would you solve that? $\endgroup$ – Boris Poris Oct 25 '20 at 3:42
  • $\begingroup$ Please create a new mathSE query, state the general question (or the specific question of 17B's, 25 other letters and 12-letter words), includce a link to this question, and indicate that your new query is a followup question. $\endgroup$ – user2661923 Oct 25 '20 at 4:08
  • $\begingroup$ Just did so, the new question is here: math.stackexchange.com/questions/3880084/… $\endgroup$ – Boris Poris Oct 25 '20 at 4:23

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