Question

Letters of the word INDIANOIL are arranged at random. Probability that the word formed

Answer 1

Total number of ways of arranging letters of the word INIDIANOIL is $\frac{9!}{3!2!}$
A) Treating INDIAN as a single object we can permute INDIAN, O, I and L in 4! ways.
$\therefore$ probability of the required event $=\frac{4!3!2!}{9!}=\frac{1}{{(}^7{C}_3\left){(}^9{C}_2\right)}$
B) We can permute OIL I, N, D, I, A, N in $\frac{7!}{2!2!}$ ways.
$\therefore$ probability of the required event $=\frac{2!2!3!2!}{7!9!}=\frac{1}{{(}^5{C}_2\left){(}^7{C}_2\right)(9!)}$
C) Fixing an I at the first place and L at the last place, we can permute the remaining letters viz. A, D, I, I, N, N, O in $\frac{7!}{2!2!}ways$.
$\therefore$ probability of the required event is $=\frac{2!2!3!2!}{7!9!}=\frac{1}{{(}^5{C}_2\left){(}^7{C}_2\right)(9!)}$
D) Vowels can be arranged at odd places viz $1^{st},3^{rd},5^{th},7^{th}$ and $9^{th}$ in $\frac{5!}{3!}ways.$
The remaining letters can be arranged at 4 even places in $\frac{4!}{2!}ways$.
$\therefore$ probability of the required event $=\frac{5!4!}{3!2!}\times \frac{3!2!}{9!}=\frac{1}{{}^9{C}_4}=\frac{1}{{}^9{C}_5}$