Question

Number of $4$ digit numbers of the form $N=abcd$ which satisfy following three conditions (i) $4000\le N<6000$ (ii) $N$ is multiple of $5$ (iii) $3\le b<c\le 6$ is equal to

• A. $12$
• B. $18$
• C. $24$
• D. $48$

Answer 1

The correct option is C $24$

$\left(i\right)4000\le N<6000\Rightarrow a=4ora=5\left(2possibilities\right)$

$\left(ii\right)Nismultipleof5\Rightarrow d=0ord=5\left(2possibilities\right)$

$\left(iii\right)3\le b<c\le 6\Rightarrow 3cases$ are possible

$\left(a\right)b=3$ then $c=4orc=5orc=6$

$\left(b\right)b=4$ then $c=5orc=6$

$\left(c\right)b=5$ then $c=6$

$=6possibilities$

$\therefore$ Total number of 4 digit numbers $=2\times 2\times 6=24$