Question

The probability that in the random arrangement of the letters of the word ‘UNIVERSITY’ the two I‘s do not come together is:

• A. $\frac{1}{5}$
• B. $\frac{4}{5}$
• C. $\frac{1}{10}$
• D. $\frac{9}{10}$

Answer 1

The correct option is B

$\frac{4}{5}$

Explanation for the correct option:

Finding the required probability:

Let $S$ be the no. of ways in which the letters of the word ‘UNIVERSITY’ can be arranged.

No. of letters in the word ‘UNIVERSITY’ $=10$

$\Rightarrow n\left(S\right)=\frac{10!}{2!}$ [As the letter I is repeated twice]

As the two I‘s can not come together, so we will first find the no. of ways in which the remaining $8$ letters can be arranged.

The remaining $8$ letters can be arranged in $8!$ ways.

So, the arrangement could be like: $\mathrm{Y}\mathrm{X}\mathrm{Y}\mathrm{X}\mathrm{Y}\mathrm{X}\mathrm{Y}\mathrm{X}\mathrm{Y}\mathrm{X}\mathrm{Y}\mathrm{X}\mathrm{Y}\mathrm{X}\mathrm{Y}\mathrm{X}\mathrm{Y}$

The places marked by X are occupied by the letters other than I, and the two I's can occupy any of the two positions marked by Y.

So, the two I's can be arranged in ${}^{9}C_2$ ways $=\frac{9!}{2!7!}=\frac{9\times 8\times 7!}{2\times 7!}=36$ $\left[{}^{n}C_r=\frac{n!}{r!\left(n-r\right)!}\right]$

Let $A$ be the no. of ways in which the letters of the word ‘UNIVERSITY’ can be arranged so that the two I‘s do not come together.

$\Rightarrow n\left(A\right)=8!\times 36$

Therefore, the required probability $P\left(A\right)$ is $=\begin{array}{rcl}& & \frac{n\left(A\right)}{n\left(S\right)}\end{array}$

$\begin{array}{rcl}\frac{n\left(A\right)}{n\left(S\right)}& =& \frac{8!\times 36}{{\displaystyle \frac{10!}{2!}}}\\ & =& 8!\times 36\times \frac{2}{10\times 9\times 8!}\\ & =& \frac{4}{5}\end{array}$

Hence, the correct answer is option B.