MHB Solve 6-Digit Combinations Question with 0-9

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Jacob needs to create a 6-digit code for his bank account using digits 0-9, with the first digit being odd and the last digit even, without repeating any digits. The calculation begins by selecting 5 options for the first digit (odd) and 5 options for the last digit (even). After choosing these, there are 8 choices for the second digit, 7 for the third, 6 for the fourth, and 5 for the fifth. The total number of combinations can be calculated as (5)(5)(8)(7)(6)(5). This approach effectively determines the number of unique 6-digit codes Jacob can create.
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"Jacob needs to create a 6-digit code for his bank account using the digits from 0 to 9. He wants the first digit to be odd and the last digit to be even. If he does not repeat any digits, how many different 6-digit codes could Jacob create?"

Would the best way to solve this be: (5)(9)(8)(7)(6)(4)?
 
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greprep said:
"Jacob needs to create a 6-digit code for his bank account using the digits from 0 to 9. He wants the first digit to be odd and the last digit to be even. If he does not repeat any digits, how many different 6-digit codes could Jacob create?"

Would the best way to solve this be: (5)(9)(8)(7)(6)(4)?

You're close, but I would begin by filling in the first and last digits first (5 choices for each), and then fill in the remaining digits:

$$N=(5)(8)(7)(6)(5)(5)=\,?$$
 
Oh, I see. So I would put in 5 first for the potential odd number (1,3,5,7,9), then 5 at the end for the potential even numbers (0,2,4,6,8), and that leaves us with 10-2 (8) choices for the second, 7 for the third, 6 for the 4th, etc?
 
greprep said:
Oh, I see. So I would put in 5 first for the potential odd number (1,3,5,7,9), then 5 at the end for the potential even numbers (0,2,4,6,8), and that leaves us with 10-2 (8) choices for the second, 7 for the third, 6 for the 4th, etc?

Yes, that's what I was suggesting. :)
 
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