Question

Words are formed using all the letters of the word INTEGRATION. A word is then picked randomly.

• A. The probability that the word starts with $E$ and ends in $N$ is $\frac{1}{55}$
• B. The probability that the word either starts with $I$ or ends in $N$ is $\frac{18}{55}$
• C. The probability that order of vowels in the word remains same is $\frac{1}{125}$
• D. The probability that the vowels and consonants alternate in the word is $\frac{1}{462}$

Answer 1

Answer: (A), (B), (D)

The probability that the vowels and consonants alternate in the word is $\frac{1}{462}$

INTEGRATION
It has $2$N, $2$I, $2$T, $1$R, $1$E, $1$G, $1$A, $1$O
Total number of words $=\frac{11!}{(2!{)}^3}$

$\to E---------N$
Number of words $=\frac{9!}{2!2!}$
Required probability $=\frac{\frac{9!}{2!2!}}{\frac{11!}{(2!{)}^3}}=\frac{1}{55}$

$\to$ Number of words starting with $I$ or ends with $N$
$=$ starts with $I$ only $+$ ends with $N$ only $-$ (starts with $I$ as well as ends with $N$)
$=\frac{10!}{2!2!}+\frac{10!}{2!2!}-\frac{9!}{2!}=\frac{9\cdot 9!}{2}$
Required probability $=\frac{\frac{9\cdot 9!}{2}}{\frac{11!}{(2!{)}^3}}=\frac{18}{55}$

$\to$ Total number of words $={}^{11}{C}_5\cdot \frac{6!}{2!2!}=\frac{11!}{5!2!2!}$
Required probability $=\frac{\frac{11!}{5!2!2!}}{\frac{11!}{(2!{)}^3}}=\frac{1}{60}$

$\to$ There are $6$ consonants and $5$ vowels.
$\therefore$ Each word starts with a consonant.
$\therefore$ Number of words $=\frac{6!}{2!2!}\times \frac{5!}{2!}$
Required probability $=\frac{\frac{6!}{2!2!}\times \frac{5!}{2!}}{\frac{11!}{(2!{)}^3}}=\frac{1}{462}$