Question

All letters of the word 'CEASE' are arranged randomly in a row, then the probability that two E are found together is

• A. $\frac{7}{5}$
• B. $\frac{3}{5}$
• C. $\frac{2}{5}$
• D. $\frac{1}{5}$

Answer 1

The correct option is C $\frac{2}{5}$

Sample space arrangement for the word ${}^{\prime }CEAS{E}^{\prime }=\frac{5!}{2!}$
Here, $E\to 2,C\to 1,A\to 1,S\to 1$
Now, consider $2E$ as one character, so that arranging for four letters $=4!ways$
$\therefore$ Required probability $=\frac{2!\times 4!}{5!}=\frac{2}{5}$