Congratulations to the following members for their correct solutions:
1) Sudharaka
Solution (from Sudharaka):
[sp]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:
[math]\frac{ \binom{14}{4} \binom{10}{4} \binom{6}{6}}{2! 2!}[/math] [/sp]