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How many 4-digit numbers have the second digit even and the fourth digit at least twice the second digit?


Please expalin/give me a hint on how to do it! Thanks!

 Apr 30, 2022
 #1
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Currently, I have 9*9*2*8 as there are 9 ways for the 1st and 3rd digit, 2 ways for the second digit, and 8 ways for the last one. This is incorrect and I do not know why so it would be nice if you could explain why im incorrect

 Apr 30, 2022
 #2
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9 possibilities for the first and third digits, and if the second digit is 2, there are 6 possibilities for the last digit, and if the second digit is 4, there are 2 possibilities for the last digit.

 

So the answer is 9*9*(6 + 2) = 648.

 Apr 30, 2022

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