Question

The letters of the word $\text{ALLAHABAD}$ are arranged at random. The probability that in the word so formed, all similar letters are found together, is

• A. $\frac{1}{63}$
• B. $\frac{16}{17}$
• C. $\frac{51}{9!}$
• D. $\frac{1}{81}$

Answer 1

The correct option is A $\frac{1}{63}$

Lets find the total number of ways in which $\text{ALLAHABAD}$ is arranged irrespective of any condition.
There are $4\text{A's}2\text{L's}1\text{B}1\text{D}1\text{H}$ in word.
Total ways of arrangement will be $\frac{9!}{4!2!}=7560$ ways

Now, ways in which all similar words are together, thus we have to consider $4\text{A's}$ as one unit, similarly $2\text{L's}$ as a unit, so number of ways will be $5!$

Thus, probability will be $\frac{5!}{7560}=\frac{1}{63}$