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

3-digit even numbers are to be formed using the given six digits, 1, 2, 3, 4, 6, and 7, without repeating the digits.

Then, units digits can be filled in 3 ways by any of the digits, 2, 4, or 6.

Since the digits cannot be repeated in the 3-digit numbers and units place is already occupied with a digit (which is even), the hundreds and tens place is to be filled by the remaining 5 digits.

Therefore, the number of ways in which hundreds and tens place can be filled with the remaining 5 digits is the permutation of 5 different digits taken 2 at a time.

Number of ways of filling hundreds and tens place = $^5P_2=\frac{5!}{(5-2)!}=\frac{5!}{3!}$
$=\frac{5\times4\times3!}{3!}=20$

Thus, by multiplication principle, the required number of 3-digit numbers is $3 \times 20 = 60$