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Question

# How many six-digit numbers can be formed from the digits 1,2,3,4,5,6, and 7, so that the digits should not repeat and the terminal digits should be even?

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Solution

## The $7⋅6⋅5⋅4⋅3⋅2$7⋅6⋅5⋅4⋅3⋅2 would be the correct answer if we didn't have the final digit being even constraint. But as we do, it makes sense to think of the number in two separate parts: the last digit, and then the first five.As you say, there are three choices for the terminal digit. But then, for the first five digits, we don't have a choice of seven digits anymore - we have the choice of six (${1,3,5,7}${1,3,5,7} or one of the unused two even digits). So the number of options for the first five is $6⋅5⋅4⋅3⋅2=720$6⋅5⋅4⋅3⋅2=720.The final answer is, therefore, $3⋅720=2160$3⋅720=2160 possible numbers.

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