Question

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

Answer 1

Here total number of digits = 6 

The unit place can be filled with any one of the digits 2, 4, 6.

So number of permutation = ³P₁ = 3!/2! = 3

Now the tens and hundred place can be filled by remaining 5 digits.

So number of permutation = ⁵P₃ = 5!/3! = (5 x 4 x 3!)/3! = 20

Hence total number of permutations = 3 x 20 = 60