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In class we are beginning discrete probability and we are being introduced to counting. He said in an example today that the word "anagram" (which has $7$ letters) can be rearranged to have $7!/3!$ possible words when we don't count the order of the a's. To me this seemed like a combination but this result reduces to $7 \cdot 6 \cdot 5 \cdot 4$ which means we are choosing $7$ letters on $4$ spots. So a bit confused if anyone could help?

N. F. Taussig
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aspookyghost20
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  • Your title is misleading. We do care about the order of the letters. However, we are only interested in distinguishable arrangements of the letters of the word anagram, so we do not care about permutations of the three as among themselves. – N. F. Taussig Oct 11 '17 at 02:11
  • What does “without order” mean? – gen-ℤ ready to perish Oct 11 '17 at 04:29
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    @N.F.Taussig, I have suggested an edit to the title that previously (incorrectly, as you pointed out) stated "the order does not matter". More than one year on, having a more accurate title seems more useful. Commenting here so your comment can be understood. – PatrickT Jan 26 '19 at 06:31
  • @user35395 That is not a correct duplicate. I saw one other post you flagged as a duplicate which was also inappropriate. Please read more carefully. – Mike Earnest Feb 16 '23 at 19:16

3 Answers3

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This can be explained as the following using multiplication principle:

  • Pick which of the seven available spaces is occupied by the n
  • Pick which of the six remaining available spaces is occupied by the g
  • $\vdots$
  • Pick which of the four remaining spaces is occupied by the m
  • All remaining three spaces will be occupied by the a's.

Applying multiplication principle, there are $7\cdot6\cdot5\cdot4$ ways to do this.

JMoravitz
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Your instructor is counting distinguishable arrangements of the word anagram.

The word anagram has seven letters, so we have seven positions to fill with $3$ as, $1$ g, $1$ m, and $1$ r. We can fill three of these seven positions with as in $\binom{7}{3}$ ways. The remaining four letters are distinct, so they can be arranged in the remaining four positions in $4!$ ways. Hence, the number of distinguishable arrangements of the word anagram is $$\binom{7}{3}4! = \frac{7!}{3!4!} \cdot 4! = \frac{7!}{3!}$$ The factor of $3!$ in the denominator represents the number of ways the three as can be permuted among themselves within a given arrangement of the letters of the word anagram without producing an arrangement distinguishable from the given arrangement.

N. F. Taussig
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I think what you notice is that the process reminds you of the number of permutations (not selections, as you wrote) of $4$ out of $7$ objects. This is indeed the case.

Now, first note that to count the number of arrangements of $n$ distinct objects, you may proceed as follows:

Partition the set in two, one containing $r$ of these $n$ objects. Then the number of arrangements is (by the multiplication principle) $$nPr\times (n-r)P(n-r)=nPr\times (n-r)!=n!.$$ This means that for each arrangement of the $r$ objects in the first part, we have $(n-r)!$ arrangements of the $n$ objects. Now, if these $n-r$ objects are indistinguishable, then the number of permutations of the $n$ (previously distinct) objects should be divided by $(n-r)!.$ This is what you noticed; and, as I have explained, hopefully you can see why it is true that the number of arrangements of $n$ objects, $n-r$ of which are indistinguishable, is the same as the number of arrangements of $r$ out of $n$ distinct objects.

Allawonder
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