# combinatorics counting words

How many di fferent 5-letter words (= strings) can be formed from the letters in CALCULUS if no letter can be used more times than it occurs in the word CALCULUS? For example, CCUUL and LSUCS would count but UUCSU would not count.

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Better if you show us how far you have gotten, where you have gotten stuck. Also, if it's homework, please use the "homework" tag. –  Gerry Myerson Jan 28 '13 at 4:52
There have been similar questions that might be helpful, e.g., here. –  Jair Taylor Jan 28 '13 at 6:41
I count only one S in CALCULUS. So LSUCS shouldn't count. –  Marc van Leeuwen Jan 28 '13 at 12:38

you have $CC,A,LL,UU,S$

Possible cases are:

$1.)$Two identical,Two identical,One distinct

$2.)$Two identical,Three distinct

$3.)$All distinct

EDIT:

For case $1.)$ ways of selection = ${3\choose2}{3\choose 1}$ and thus total such words= ${3\choose2}{3\choose 1}\frac{5!}{2!2!}$

For case $2.)$ ways of selection = ${3\choose1}{4\choose 3}$ and thus total such words= ${3\choose1}{4\choose 3}\frac{5!}{2!}$

For case $3.)$ ways of selection = ${5\choose5}$ and thus total such words= ${5\choose5}5!$

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HINT: The word CALCULUS has three letters that occur twice each (C, L, U) and two that appear once each (A, S). To make a $5$-letter word clearly requires using at least $3$ different letters.

• If we use $5$ different letters, we must use exactly one each of A, C, L, S, and U; this is the only possible set of $5$ letters that we can use. It can be permuted in $5!$ ways, for a total of $5!=120$ words.

• If we use $3$ different letters, at least $2$ of them must come from the set $\{\text{C, L, U}\}$. If all three come from that set, there are $3$ ways to choose which of the three letters occurs only once, and $\frac{5!}{2!2!}=30$ distinct ways to permute a given set of two doubletons and a singleton, for a total of $3\cdot 30=90$ words. If just two of the three letters come from the set of doubletons, there are $\binom32=3$ ways to choose which $2$ they are, and there are $\binom21=2$ ways to choose which of the singletons is used. Once again there are $30$ distinct ways to permute a given set of two doubletons and a singleton, so this subcase produces another $3\cdot2\cdot30=180$ words. The case as a whole therefore includes $90+180=270$ words.

The remaining case is the one in which we use $4$ different letters. Can you work it out, now that you’ve seen the other two cases?

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this is a fairly straight forward combinatorial exercise.

One method usually suggested is to label the repeated letters. So, instead of having 2 Cs, you can call them $C_1$ and $C_2$. From there, you can then check for double counting.

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