Problem of the week #23 - September 3rd, 2012

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Jameson

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MATH HELP BOARDS

is three words. The first word contains 4 letters, the second 4 letters and the third 6 letters. How many different ways can you group these letters into groups of 4, 4 and 6?

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Jameson

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Congratulations to the following members for their correct solutions:

1) Sudharaka

Solution (from Sudharaka):

This problem is unfortunately ambiguous. I didn't specify if permutations of the same group count or not. If they do count, then here is Sudharaka's solution.
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Altogether there are 14 letters and there are two A's, two H's. All the other letters occur only once. If there are 14 different letters then there are $$14!$$ different permutations. Let $$N$$ be the number of permutations considering the repetitions of letters. For each permutation the two A's can be interchanged. Similarly the two H's can also be interchanged. Therefore,

$2!\times 2!\times N=14!$

$\Rightarrow N=\frac{14!}{2!\times 2!}=21794572800$

Finally each permutation above can be grouped into three groups of 4, 4, 6 letters. Hence the number of ways that the letters can be formed into the three groups of 4, 4, 6 letters is also, $$21794572800$$.
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If permutations of the same group don't count, then the solution is:

$$\displaystyle \frac{ \binom{14}{4} \binom{10}{4} \binom{6}{6}}{2! 2!}$$

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