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?

Asked by Pragya Singh | 1 year ago |  84

##### Solution :-

From the digits 1,2 ,3 ,4 and 5, 4-digit numbers can be formed. There are permutations of 5 different things taken 4 at a time. Thus, number of 4 digit numbers

$$^5P_4=\dfrac{5!}{(5-4)!}=\dfrac{5!}{1!}$$

= 1×2×3×4×5=120

Out of 1, 2, 3, 4, 5, we know that even numbers end either by 2 or 4. Thus, ways in which units place can be filled is 2. Since, repetition is not allowed, units place is already occupied by a digit and remaining vacant places can be filled by remaining 4 digits. Thus, number of ways in which remaining places can be filled =

$$^4P_3=\dfrac{4!}{(4-3)!}=\dfrac{4!}{1!}$$

= 4×3×2×1= 24

Therefore, by multiplication principle, number of even numbers  $$24\times 2=48$$

Answered by Abhisek | 1 year ago

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