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# counting problem

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35
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How many 4-digit numbers have the second digit even and the fourth digit is at least the second digit? (Here, digits are read from the left, so the first digit is the leftmost digit, and so on.)

Jun 8, 2022

#1
+1746
+1

Note that regardless of the second digit, there will always be 90 ways to choose the first and third digits (9 for the first, and 10 for the third).

If the second digit is a 0, there are 10 choices for the final digit.

If the second digit is a 2, there are 8 choices for the final digit.

If the second digit is a 4, there are 6 choices for the final digit.

If the second digit is a 6, there are 4 choices for the final digit.

If the second digit is an 8, there are 2 choices for the final digit.

This means the number of 4 digit numbers is: $$9 \times 10 \times (10 + 8 + 6 + 4 + 2) = \color{brown}\boxed{2,700}$$

Jun 8, 2022
#2
+117487
+1

nice logic builderboi but you did make a small error.

You can fix it if you want but you do not need to fix it,

Guest can find it for him/her self.

Jun 9, 2022