How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, and 7 if no digit is repeated?

Question: How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, and 7 if no digit is repeated?

Solution:
Given digits: 1, 2, 3, 4, 6, 7
Number of digits = 6

Number of possible digits at unit’s place = 3 (2, 4 and 6)
⇒ Number of permutations = ³P₁ = 3

When one of the digits is taken in units’ place, then the number of possible digits available = 5
⇒ Number of permutations = ⁵P₂ = 5!/(5 - 2)!
= 5!/3!
= 120/6
= 20

∴ The total number of permutations = 3 × 20 = 60.