Question

The first $12$letters of the English alphabet are to be written at random so that there are exactly$4$ letters between $A$and$B$. the number of ways this can be done is?

• A. $7\times 10!$
• B. $2\times 10!$
• C. $21\times 10!$
• D. $14\times 10!$

Answer 1

The correct option is D

$14\times 10!$

The explanation for the correct answer.

The number of ways to arrange $A$and$B$ is,

$A_1{B}_6$ , $A_2{B}_7$ ,$A_3{B}_8$$...,$ $A_7{B}_{12}$

The number of ways the arrangement of remaining letters in $10$ places can be done in $10!$ways.

So, the total number of arrangements $=2x7x10!=14x10!$

Hence. option(D) is the correct answer.