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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?

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?

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4-digit numbers that can be formed using the digits 1, 2, 3, 4,5.

Therefore, there will be as many 4-digit numbers as there are permutations of 5 different digits taken 4 at a time.

Therefore, the required number of 4-digit numbers $=^{5}P_4$

$=\frac{5!}{(5-4)!}$

$=\frac{5!}{1!}$

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

4-digit even numbers can be made using the digits  1, 2, 3, 4, 5 if no digit is repeated.

The unit place can be filled in 2 ways by any digits from 2 or 4.

The digit cannot be repeated in 4-digit numbers and the unit place is occupied with a digit(2 or 4).

Thousands, hundreds, tens place can be filled by remaining any 4 digits.

Therefore, there will be as many 3-digit numbers as there are permutations of 4 different digits taken 3 at a time.

Therefore, the required number of 3-digit numbers $=^{4}P_3$

$=\frac{4!}{(4-3)!}$

$=\frac{4!}{1!}$

$=4\times 3\times 2\times 1=24$

Thus, by multiplication principle, required 4 -digit numbers is  $2\times 24=48$

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