SOLUTION: How many different 4-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6, 8, and 9 if no repetition of digits is allowed.

Question 1173645: How many different 4-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6, 8, and 9 if no repetition of digits is allowed.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52776) (Show Source): You can put this solution on YOUR website!
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Was solved a week ago under this link

https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.1173334.html

https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.1173334.html

Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!

The only restriction on the number to be formed, other than the requirement that the digits be all different, is that the number must be even.

That means the units digit must be even; the other digits can be any digits.

Since the only restriction is on the units digit, choose it first. Among the given digits, 4 of them are even, so there are 4 choices for the units digit.

After that, there are 8 digits left to choose from; and you can choose the tens, hundreds, and thousands digits in any order you want.

Because of the requirement that the digits all be different, there are 8 choices for the next digit you choose, 7 choices for the one after that, and 6 choices for the last.

So the total number of 4-digit even numbers that can be formed from the given digits is 4*8*7*6 = 1344

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