How many distinct n-letter "words" can be formed from a set of k letters where some of the letters are repeated?



How many 6-letter words can be formed from the letters: ABBCCC?

This is elementary. There are 6! arrangements counting repeats. Then we just divide by 2!3! to account for the repeats caused by the 2!3! identical arrangements of the Bs and Cs

How many 5-letter words can be formed from the letters: ABBCCC?

Excluding A we have $\frac{5!}{2!3!}$

Excluding either B $\frac{5!}{3!}$

Excluding any of the Cs $\frac{5!}{2!2!}$

Then take the sum.

How many 4-letter words can be formed from the letters: ABBCCC?

At this point I find it difficult to procede without far too many cases.

Is there a general approach?

  • $\begingroup$ Interesting question, perhaps difficult to give a nice answer to. In the particular instance, we remove AB or AC or BB or CC or BC, so the number of cases is manageable. $\endgroup$ – André Nicolas Nov 16 '12 at 22:11
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    $\begingroup$ There's a nice exponential generating function for this problem. See this question. $\endgroup$ – Jair Taylor Nov 17 '12 at 20:35

There is a fairly systematic way to work this type of question. You were specifically wondering about:

How many 4-letter words can be formed from the letters: ABBCCC?

First, you write the 'source partition' for your word:

[3,2,1]  <-- this is the `source partition`

Note that the source partition provides for 1 triple, 1 double, and 1 single. Corresponding to that, you write every possible 'target partition' of 4-letters.

[3,1]    requests one triple and 1 single
[2,2]    requests 2 doubles
[2,1,1]  requests one double and 2 singles

For each 'target partition', you write an expression that gives the number of ways that the given target type can occur. An example of the target type [3,1] is:

CCBC                                   # type [3,1]

The expression for that type is:

nCr(1,1)*nCr(2,1) * fac(4)/fac(3)

'nCr( n, r)' is the function for 'combinations', which gives the number of ways you can make a unique selection from n distinct things, taking r at a time. 'fac( n)' is the function for the factorial of n.

Note that source [3,2,1] provides 1 triple, and target [3,1] requests 1 triple. Hence nCr(1,1). After the triple is used up, the source [3,2,1] can provide 2 singles, and the target [3,1] requests 1 single. Hence nCr(2,1) The call to fac(4), in each expression, always corresponds to the 4-letter word. Any division, if it occurs in an expression, corresponds to each multiple request of the corresponding target partition. That's all there is to the method, but it isn't always easy to get every last detail correct. The entire computation, as I did it in the programming language Python, follows:

# The `source partition` for the word 'ABBCCC' is:
#                 [3,2,1]
#                                                # target
w = 0                                            # ------
w += nCr(1,1)*nCr(2,1) * fac(4)/fac(3)           # [3,1]
w += nCr(2,2)          * fac(4)/(fac(2)*fac(2))  # [2,2]
w += nCr(2,1)*nCr(2,2) * fac(4)/fac(2)           # [2,1,1]
# The answer is 38

Let's say you want the word of letters to be $M$-long, with $N$ choices for letters in each space on the word. Then draw something like this:

enter image description here

to enumerate all possible words. In the picture $M=3$ and $N=2$. Let's say you have a deficit in $A$s of magnitude $D_A$ from being able to make a full, $M$-long word of $A$s.

First, assuming none of the deficits overlap (that is, there are no words that violate 2 or more restrictions in the number of letters), find the number of nodes that are less than or equal to $M-D_A$ away from $(A,A,...A)$. This turns out to be $\sum_{k=0}^{k=M-D_A}\binom{M}{k}{(N-1)}^{M-k}$.

For the proof of this, think of a 3-letter word with choices of letter $A$, $B$ or $C$. There are $\binom{3}{0}2^0$ ways of making words with 3 $A$s, $\binom{3}{1}2^1$ ways of making a word with 2 $A$s, etc.

So evidently you need to subtract each sum taking into account the restrictions of each letter from the total number of unrestricted words.

However, it's much more complicated when the sum of two deficits is more than $M$ so that because of the overlap the subtraction will have been too great. Basically the $(N-1)$ in the sum will be some lower number for higher values of $k$ when other letter's restrictions are overstepped, but I can't see a nice formula for it for a given $D_B, D_C...$.


It is solved here. In general, you have to account for the duplicates as detailed in the link.


where the denominator denotes factorial for indistinguishable or repeated letters.

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    $\begingroup$ Thanks, but, everything at that link is equivalent to my first example only. That formula will not work when n < k. $\endgroup$ – futurebird Nov 16 '12 at 22:02

I'm pretty sure it's just 6!/(2!*2!*3!) which is equal to 30. Where the first 2! shortens the word to 4 letters, same with the 5 letter one it would just me 1!, but omitted cause its just 1. I even went and counted the combinations (in image). What I counted. If I'm totally missing something please tell.

  • $\begingroup$ Writing things out manually by brute force should only be a last resort. You did miss things, including CBCA, CBCB, CBCC and others $\endgroup$ – JMoravitz Jan 9 '20 at 19:50

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