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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?

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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?

Sum
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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 =  

5P4 = `(5!)/((5 - 4)!) = (5!)/(1!)`

= 1x 2 x 3 x 4 x 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 are 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

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Chapter 6: Permutations and Combinations - EXERCISE 6.3 [Page 114]

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NCERT Mathematics [English] Class 11
Chapter 6 Permutations and Combinations
EXERCISE 6.3 | Q 4. | Page 114

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