Given,
Digits which can be used to make numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9
The number of these digits are 10
To find : Total number of three-digit numbers with no digit repeated
Formula used :
Number of arrangements of n things taken r at a time = P(n, r)
P(n, r) = \(\frac{n!}{(n-r)!}\)
∴ The total number of ways
= the number of arrangements of 10 things taken 3 at a time
= P(10, 3)
= 10 × 9 × 8
= 720
But in these 720 numbers we also included those numbers which are starting from 0 like 023 or 056 etc.
Being starting from 0, these are actually two-digit numbers.
So, we need to subtract these numbers.
To find these numbers, fix the position of 0 at hundred’s place.
Remaining numbers = 9 (1, 2, 3, 4, 5, 6, 7, 8 or 9)
Arrange these 9 numbers in remaining 2 places.
Formula used :
Number of arrangements of n things taken r at a time = P(n, r)
P(n,r) = \(\frac{n!}{(n-r)!}\)
∴ The total numbers which are starting from 0 are
= the number of arrangements of 9 things taken 2 at a time
= P(9, 2)
= 9 × 8
= 72
Hence,
Total number of three-digit numbers with no digit repeated are, 720 – 72
= 648
