The $7$ digits $1, 2, 3, 5, 6, 8, 9$ are to be used to make $5$ digit numbers with each being used not more than once in a number. How many numbers can be made which are more than 60 000 AND even? I got something like $3\cdot 3\cdot 5P_3 = 540$. But I think it's wrong because there are numbers that we can choose which are both even and $\geq 6$.
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Hint: Choose the first digit first, dividing into cases based on whether it is even or odd. How many ways for each? Then choose the ones digit, noting that it has to be even. How many ways? Then fill in the rest.
Ross Millikan
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Well, the first digit can be either 6, 8 or 9; that's 3. And the last digit can be either 2, 6 or 8; that's also 3. – mbkoverflow Oct 08 '17 at 14:58
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The rest 3 digits can be any of the remaining 5. But what if a number has a 6 in the beginning. That 6 can't be used again in the last digit, right? – mbkoverflow Oct 08 '17 at 14:59
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That is why I suggested you divide into cases based on the parity of the first digit. If the first digit is even you only have two choices for the ones digit. If it is odd you have three. – Ross Millikan Oct 08 '17 at 15:01
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Okay so now I got: – mbkoverflow Oct 08 '17 at 15:31
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1225P3 + 135P3 = 240 + 180 = 420. – mbkoverflow Oct 08 '17 at 15:32
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That is right... – Ross Millikan Oct 08 '17 at 18:59
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Thank you so much for the help. <3 – mbkoverflow Oct 08 '17 at 19:42