# 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?

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?

Solution:

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 $={ }^{5} \mathrm{P}_{4}=\frac{5 !}{(5-4) !}=\frac{5 !}{1 !}$

$=1 \times 2 \times 3 \times 4 \times 5=120$

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.

Number of ways of filling the remaining places $={ }^{4} \mathrm{P}_{3}=\frac{4 !}{(4-3) !}=\frac{4 !}{1 !}$

= 4 × 3 × 2 × 1 = 24

Thus, by multiplication principle, the required number of even numbers is

24 × 2 = 48