Aspire Faculty ID #15475 · Topic: MAH MCA CET 2025 (Shift 1) · Just now
MAH MCA CET 2025 (Shift 1)

Find the number of arrangements of the letters of the word $,\text{INDEPENDENCE},$ such that
(i) the words start with $I$ and end with $P$.
(ii) all the vowels never occur together.

Solution

Letters in INDEPENDENCE: $E^4,\ N^3,\ D^2,\ I,\ P,\ C$ (total $12$).

(i) Fix $I$ at start and $P$ at end. Arrange the remaining $10$ symbols: $E^4,\ N^3,\ D^2,\ C$.
Number $= \dfrac{10!}{4!,3!,2!}=12600$.

(ii) Total arrangements $= \dfrac{12!}{4!,3!,2!}=1663200$.
“All vowels together”: treat $(\text{vowels}=E^4I)$ as one block. Then items are
${\text{block}, N^3, D^2, C, P}$ → $8$ items with repeats $N^3, D^2$.
Arrangements of items $= \dfrac{8!}{3!,2!}=3360$.
Internal vowel-block arrangements $= \dfrac{5!}{4!}=5$.
Together $= 3360\times5=16800$.
Hence “never together” $=1663200-16800=1646400$.
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