Question

The letters of the word "FLOWER" are taken $4$ at a time and arranged in all possible ways. The number of arrangements which begin with ${}^{\prime }{F}^{\prime }$ and end with ${}^{\prime }{R}^{\prime }$ is

• A. $20$
• B. $18$
• C. $12$
• D. $14$

Answer 1

The correct option is C $12$

Since $F$ and $R$ occupy first and the last position, we can select any two letters out of the remaining four.

These can be done in

${}^4{C}_2$ ways

i.e. ${}^n{C}_r=\frac{n!}{(n-r)!r!}$

$=\frac{4!}{(4-2)!2!}$

$=\frac{4!}{2!2!}=\frac{4\times 3\times 2\times 1}{2!2!}=6$

These $2$ selected letters can be arranged themselves in $2$ ways.

Total number of ways $=6\times 2=12$

Hence, the answer is option C.