How many 3 - digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed?

Concept:

Permutation: Permutation is defined as an arrangement of r things that can be done out of total n things. This is denoted by 

\({\;^n}{P_r} = \frac{{n!}}{{\left( {n - r} \right)!}}\)

Calculation:

Here order matters for example 123 and 132 are two different numbers. Therefore, there will be as many 3 digit numbers as there are permutations

of 9 different digits taken 3 at a time.

Therefore, the required 3 digit numbers 

⇒ In 3^rd place required number = 9

⇒ in 2^nd place required number = 8

⇒ In 1^st place required number = 7

So, 3 - digit number form by 

⇒ 9 × 8 × 7

⇒ 504

Additional Information

Combination: The number of selections of r objects from the given n objects is denoted by  

ⁿC_r = \(\rm \frac{n!}{r! (n - r)!}\)