Class 11

Math

Algebra

Permutations and Combinations

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

- $62$
- $60$
- $64$
- $65$

Solution :-

number lying between 100 and 1000

formed by 3 digits.

and every digit have 5 option (1,2,3,4,5)

to select a number

But repetion is not allowed

so number = $_{5}P_{3}$

$=(5−3)!5! $

$=2!5! $

$=60m.$