Mathematics - Permutation Combination Question with Solution | TestHub

(A) If ENDEA is fixed word, then assume this as a single letter. Total number of letters $=5$ and total number of arrangements $=5$ ! . (B) If $E$ is at first and last places, then total number of permutations $$\frac{7!}{2!}=21\times 5!$$ (C) If $\mathrm{D},\mathrm{L},\mathrm{N}$ are not in last five positions $$\leftarrow \mathrm{D},\mathrm{L},\mathrm{N},\mathrm{N}\to \leftarrow \text{ E, E, E, A, O }\to$$ Total number of permutations $=\frac{4!}{2!}\times \frac{5!}{3!}=2\times 5$ ! . (D) Total number of odd positions $=5$ Permutations of AEEEO are $\frac{5!}{3!}$ Total number of even positions $=4$ Number of permutations of $\mathrm{N},\mathrm{N},\mathrm{D},\mathrm{L}=\frac{4!}{2!}$ Hence, total number of permutations $=\frac{5!}{3!}\times \frac{4!}{2!}=2\times 5!$.