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 = ${}^3{P}_1=\frac{3!}{2!}=3$

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

So number of permutations =${}^5{P}_2=\frac{5!}{3!}=\frac{5\times 4\times 3!}{3!}=20$

Hence total number of permutations = $3\times 20=60.$