Find the number of permutations of the letters of the word ALLAHABAD.

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5880

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7560

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Solution

The correct option is C
7560

In the word ALLAHABAD, there are 9 letters, in which A appears 4 times, L appears 2 times and H,B,D each appears once.

Therefore, the required number of arrangements $\frac{9!}{(4! \times 2!)} = 7560$

Explanation:

A appears 4 times, L appears 2 times

Temporarily, let us treat all these repeated letters different and name them as $A_{1}, A_{2}, A_{3}, A_{4}, L_{1}, L_{2}$ , The number of permutations of 9 different letters, in this case, taken all at a time is 9!.

One such permutation is $L_{1} A_{1} H A_{2} D A_{3} B A_{4} L_{2}$ .
Here if $A_{1}, A_{2}, A_{3}, A_{4}$ are not same and $L_{1}, L_{2}$ are not same, then $A_{1}, A_{2}, A_{3}, A_{4}$ can be arranged in 4! ways and $L_{1}, L_{2}$ can be arranged in 2! ways.

Therefore, $4! \times 2!$ permutations will be just the same permutation corresponding to this chosen permutation $L_{1} A_{1} H A_{2} D A_{3} B A_{4} L_{2}$ . Hence the total number of different permutations will be $\frac{9!}{(4! \times 2!)} = 7560$

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