NCERT Solutions
Class 11 Maths
Permutations and Combinations

Ex.7.3 Q.4
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?
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 = 5P4 = 5! ÷ (5 – 4)!
= 5! ÷ 1!
= 5 × 4 × 3 × 2 × 1
= 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 = 4P3 = 4! ÷ (4 – 3)!
= 4! ÷ 1!
= 4 × 3 × 2 × 1
= 24
Thus, by multiplication principle, the required number of even numbers is 24 × 2 = 48