In how many different ways can the letters of the word ALLAHABAD be arranged?

### Concept & Permutations with Identical Items When calculating arrangements of words with repeated letters, divide the total factorial of the letter count by the factorials of the frequencies of the repeating letters. $$ \text{Arrangements} = \frac{n!}{p_1! \times p_2! \dots} $$ ### Step-by-Step Solution 1. Count the total letters in "ALLAHABAD": $n = 9$. 2. Tally the frequencies of the letters: - A: 4 - L: 2 - H: 1 - B: 1 - D: 1 3. Substitute these values into the permutation formula: $$ \frac{9!}{4! \times 2!} $$ 4. Expand and cancel terms: $$ \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4! \times 2} = \frac{9 \times 8 \times 7 \times 6 \times 5}{2} $$ 5. Calculate the remaining product: $$ 9 \times 4 \times 7 \times 6 \times 5 = 36 \times 210 = 7560 $$ ### Exam Strategy & Shortcut Stop expanding the numerator as soon as it matches the largest factorial in the denominator. Here, stop at $4!$ so it immediately cancels out with the $4!$ below, saving massive amounts of computation time. ### Common Pitfall A frequent error is dividing by just the number of identical letters (e.g., dividing by 4 and 2) rather than their factorials ($4!$ and $2!$). ### Final Answer Therefore, the correct answer is **7560**.