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How many permutations of "committee" exist where is must not end in an 'e' ?

I've been trying to figure out a possible angle of attack on this question.

I've tried to say instead, "how many possible ways are there that do end in an e?". However, this is causing me problems as there are letter repetitions.

Where should I start?

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4 Answers 4

up vote 3 down vote accepted

Answer:

Total number of ways "COMMITTEE" could be formed is $$\frac{9!}{2!2!2!}$$

Now fix "E" in the last position and compute the different number of ways to permute the rest. This is

$$\frac{8!}{2!2!}$$

Now the total number of ways "COMMITTEE" could be formed in such a way that it does not end with an "E" is

$$\frac{9!}{2!2!2!} - \frac{8!}{2!2!}$$

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okay great, all of the previous answers are good but yours made the most sense to me. If i am interpreting you correctly, then to "fix" e to the end, are you essentially saying you are making an 8 letter word "committe" instead? –  ylun.ca Feb 18 at 2:01
    
Yes, you are making a eight letter word committe which is different from a 9 letter word not to end in "e" –  satish ramanathan Feb 18 at 2:05
    
Both you and Andre's answers are great but yours is a more simple process to follow. Thank you. –  ylun.ca Feb 18 at 2:21
    
You are welcome –  satish ramanathan Feb 18 at 2:22

This is problem 3 of chapter 1 from Lászlo Lovász book Combinatorial Problems and Excercises. He uses the word caharacterization. Here I copy what he says about the word characterization.

There are 16! permutations of the word CHARACTERIZATION however not all of these give new words; in fact,in any permutation, if we exchange the three $A$'s,the two $C$'s,the two $R$'s,the two $I$'s or the two $T$'s we get the same word. Thus for any permutation, there are $3!\cdot2\cdot2\cdot2\cdot=96$ permutations which give the same word, so the result is $\frac{16!}{96}$

In General, if there are $k_A$ $A$'s, $k_b$ $B$'s etc. then the result is

$\dfrac{k_A+k_b+\dots}{k_A!k_B!\dots}$

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There are $9$ slots to be filled. An e cannot be at the end, so we can pick the pair of locations occupied by the e's in $\binom{8}{2}$ ways. For each of these, the locations of the m's can be picked in $\binom{7}{2}$ ways. Now the locations of the t's can be chosen in $\binom{5}{2}$ ways. The remaining $3$ slots can be filled in $3!$ ways. This gives a total of $\binom{8}{2}\binom{7}{2}\binom{5}{2}3!$ words.

Remark: It is very useful to place the e's first. The order in which we consider the rest does not matter. For example, after the e's are placed, the c can be placed in $\binom{7}{1}$ ways. For each of these, the t's can be placed in $\binom{6}{2}$ ways. and so on.

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regarding you remark: essentially then, we are making an 8 letter word with "committe" instead? –  ylun.ca Feb 18 at 2:03
    
I had a "typo" in the remark. We place the e's. That leaves a $7$-letter word to be made from committ. –  André Nicolas Feb 18 at 2:08
    
@Andre, our answers seem to differ, Am I making an error –  satish ramanathan Feb 18 at 2:08
    
where am I missing is it because we have two e's –  satish ramanathan Feb 18 at 2:11
    
Compute, they do not differ. Mine is $\binom{8}{2}\binom{7}{2}\binom{5}{2}3!$, which is $35280$. I think your $\frac{9!}{8}-\frac{8!}{4}$ is the same. –  André Nicolas Feb 18 at 2:15

$committee=c^1*o^1*m^2*i^1*t^2+e^2$

For any word how many permutations give the same word? The letters $t$ can be switched (2 ways), the letters $m$ can be switched (2 ways) and the letters $e$ can also be switched (2 ways) therefore for any word there are $2\cdot2\cdot2=8$ permutations giving the same word.

Thus there are $\dfrac {8!}{8}=7!$ words.

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committee*..... –  ylun.ca Feb 18 at 1:57
1  
Fixed, sorry for inconvenience –  Bananarama Feb 18 at 2:00

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