Question

How many numbers between 100 and 1000 can be formed without repetition using the digits from 1 to 9?

• A. $\frac{9!}{3!}$
• B. $\frac{9!}{(9-3)!}$
• C. $3!$
• D. None of the above

Answer 1

The correct option is B $\frac{9!}{(9-3)!}$

All numbers between $100$ and $1000$ are of three digit.

So, out of all the three digit numbers we have to find numbers having all the digits different (i.e without repetition) by using the digits from $1$ to $9$.

It is similar to filling of $3$ vacant place using $9$ different things without repetition.


Once place can be filled in $9$ different ways.
ten's place can be filled in $8$ different ways.
And, hundredth place can be filled in $7$ different ways.

So, once "and" ten's "and" hundredth places can be filled in $9\times 8\times 7=504.$

This can also be solved as arrangement of $9$ different things taken $3$ at a time. Which is equal to ${}^9{P}_r$.

$\Rightarrow {}^9{P}_3=\frac{9!}{(9-3)!}=\frac{9!}{6!}$

$\Rightarrow {}^9{P}_3=\frac{9\times 8\times 7\times 6!}{6!}$

$\Rightarrow {}^9{P}_3=9\times 8\times 7=504.$