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I encountered this problem:

How many $4$-digit numbers are there that do not have $5$ and have $7$ in the hundreds position? Digits cannot be repeated.

My thinking: first, let's count numbers that do not have $0$. It's going to be $7\cdot 6 \cdot 5$ arrangements. Now, let's count numbers that have $0$ in the tens position. We have two positions open so there are $7 \cdot 6$ arrangements. We get the same number of arrangements for numbers that end with $0$. The total number of arrangements is $7\cdot 6 \cdot 5+ 2\cdot 7 \cdot 6=7\cdot 7 \cdot 6=294$.

However, my friend argues that there are $8$ choices for the last digit, $7$ choices for the tens digits and $6$ choices for the first digit so the total number should be $8\cdot 7 \cdot 6=336$.

I am asking who is right not to prove my friend wrong but to find the truth. Thanks for listening!

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294 is correct.

First digit (most significant), can not be 7,0,5, so 7 choices; 3rd digit, cannot be 7,5, and the one chosen for 1st, so 7 choices; 4th digit, 6 choices;

$7\times7\times6=294$

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    $\begingroup$ Welcome to the site. Great first answer! $\endgroup$ – B. Mehta Apr 14 '18 at 23:17
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You are right and your friend is wrong. If you choose the units digit and then choose the tens digit (and are forced to use $7$ for the hundreds digit) then the number of choices for the thousands digit not always $6$: it is either $5$ or $6$, depending on if you've used $0$ yet or not.

For example, there are $5$ numbers satisfying the condition which have the form $x789$ (since $x$ can be any of $1,2,3,4,6$) but $6$ numbers satisfying the condition which have the form $x780$ (since $x$ can be any of $1,2,3,4,6,9$).

All this tells us is that the number of solutions is between $8\cdot7\cdot5 = 280$ and $8\cdot7\cdot6 = 336$, and for a more precise answer we do need to look at whether $0$ gets used.

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I think the key disagreement here is that your friend is thinking that $0$ is allowed to be the first digit (say, for a 4 digit passcode), while you are not. If $0$ can be the first digit, your friend is correct, but if not (as other posters explain), you are correct.

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You don't have to choice the positions in order so don't worry about $0$s.

Do the Thousandth place first. It can not be $0$ and it can not be $7$ or $5$. So there are $7$ options. Then do either the tens or the ones, it doesn't matter which. It can be $0$ but it can't be $7$ or $5$ and it can't be what the thousandth place was. That is $7$ options as well. And the final position can't be $7$ or $5$ or either of the previous two positions. so that is $6$ options.

So there are $7*7*6 = 294$.

You were correct but you didn't have to work so hard.

Your freind is wrong in that $0$ is not an option for the first digit.

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