Question

How many numbers lying between $100$ and $1000$ can be formed with the digits $0,1,2,3,4,5,$ if the repetition of the digits is not allowed?

Answer 1

Step$:1$ Finding total $3-$digit numbers.

Total digit $0,1,2,3,4,5$
$\therefore n=6$ and $r=3$
$\therefore$ Total number of $3$ digits number
$={}^n{P}_r={}^6{P}_3$
$=\frac{6!}{(6-3)!}$
$=\frac{6\times 5\times 4\times 3!}{3!}$
$=6\times 5\times 4$
$=120$

Step 2:
Finding $3-$ digit numbers that start with $0.$
Total digit $n=5$ and $r=2$
$\therefore$Total number of $3-$digit numbers that start with $0.$
$={}^5{P}_2=\frac{5!}{(5-2)!}$
$=\frac{5\times 4\times 3!}{3!}=5\times 4=20$

Step $3:$
finding required 3-digit numbers.

$\therefore$ Required number $=$ Total $3-$digit numbers-$3$ digit numbers which start with $0$
$=120-20=100$