Question

All possible four digit numbers are formed using the digits 0, 1, 2, 3 so that no number has repeated digits. The number of even numbers among them is.

• A. 9
• B. 18
• C. 10
• D. None of these

Answer 1

Answer: (C)

10

Answer 2

In forming even numbers, the position on the right can be filled either 0 or 2. When 0 is filled, the remaining positions can be filled in $3\mspace{6mu}!$ ways and when 2 is filled, the position on the left can be filled in 2 ways (0 cannot be used) and the middle two positions in $2\mspace{6mu}!$ ways (0 can be used). Therefore the number of even numbers formed $= 3\mspace{6mu}! + 2(2\mspace{6mu}!) = 10$.