Question

Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?

Answer 1

4-digit numbers are to be formed using the digits, 1, 2, 3, 4, and 5.

There will be as many 4-digit numbers as there are permutations of 5 different digits taken 4 at a time. Therefore, required number of 4 digit numbers =$^5P_4=\frac{5!}{(5-4)!}=\frac{5!}{1!}$
$=1\times2\times3\times4\times5=120$

Among the 4-digit numbers formed by using the digits, 1, 2, 3, 4, 5, even numbers end with either 2 or 4. The number of ways in which units place is filled with digits is 2.

Since the digits are not repeated and the units place is already occupied with a digit (which is even), the remaining places are to be filled by the remaining 4 digits.

Therefore, the number of ways in which the remaining places can be filled is the permutation of 4 different digits taken 3 at a time.
=$^4P_3=\frac{4!}{(4-3)!}=\frac{4!}{1!}$
Number of ways of filling the remaining places
$= 4\times 3 \times 2 \times 1 = 24$

Thus, by multiplication principle, the required number of even numbers is $24 \times 2 = 48$