Question

A license plate begins with three letters. If the possible letters are $A,B,C,D,E,F,G$ and $H,$ how many different permutations of these letters can be made if no letter is used more than once?

• A. $336$
• B. $8^3$
• C. $3!$
• D. $8!$

Answer 1

The correct option is A $336$

The problem involves $8$ things $(A,B,C,D,E,F,G,H)$ taken $3$ at a time.

$\therefore$ Required number of permutations
$=P(8,3)=\frac{8!}{(8-3)!}=\frac{8\times 7\times 6\times 5!}{5!}=8\times 7\times 6=336$