Discussion
The number of arrangements of the letters of the word INDEPENDENCE such that the first letter is I and the last letter is P, is
Answer: D
The number of arrangements is 12600.
INDEPENDENCE has 12 letters: E four times, N three times, D twice, and I, P, C once each. Fixing I first and P last leaves 10 letters, N, N, N, D, D, E, E, E, E, C, to arrange in the middle:
\[ \begin{aligned} \frac{10!}{3!\,2!\,4!} &= \frac{3628800}{6 \times 2 \times 24} \\ &= \frac{3628800}{288} = 12600 \end{aligned} \]
Each repeated letter divides out the orderings of identical copies. The other options lie close to 12600, so the division has to be done exactly; none of them equals \(\frac{10!}{3!\,2!\,4!}\).
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