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Find the number of three-digit numbers in which exactly one digit $3$ is used?

The number is of one of the forms $$\_\text{ }\_\text{ }3 \\ \_\text{ } 3 \text{ } \_ \\ 3\text{ }\_ \text{ }\_$$ There are $V_9^2-V_8^1=9\times8-8=64$ possibilities for each of the first $2$ forms, and $V_9^2$ for the third. This makes $2\times64+72=200$ possibilities in total. The given answer in my book is $225$. What am I missing?

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    $\begingroup$ We have $9$ digits other than $3$, but $0$ can't be the first digit so for both the first and second we have $8\times 9=72$ possibilities each. For the last one we have $9\times9=81$ possibilities. $\endgroup$ – Shubham Johri Apr 3 at 9:27
  • $\begingroup$ Are you open to alternative ways to approach this question, or do you just want to know why your steps were wrong? $\endgroup$ – Toby Mak Apr 3 at 9:28
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    $\begingroup$ @TobyMak, I am open to alternative ways to approach this question! But of course I want to know why my steps are wrong. $\endgroup$ – Medi Apr 3 at 9:29
  • $\begingroup$ Never mind, Light Yagami beat me to it. $\endgroup$ – Toby Mak Apr 3 at 9:30
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    $\begingroup$ You have probably ignored that digits other than $3$ can be repeated $\endgroup$ – Shubham Johri Apr 3 at 9:30
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First, fill all places. $9\times 9\times 3=243$ possibilities. Now remove $0's$ which appear at the front, whose only possibilities are in the first two cases which are $9+9=18$. Hence $243-18=225$.

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  • $\begingroup$ Aren't all possibilities to choose 3 from 10 digits $V_{10}^3=10\times9\times8$? $\endgroup$ – Medi Apr 3 at 9:31
  • $\begingroup$ @Medi 3 cannot be used again and rest all digits can be repeated. $\endgroup$ – Harry Potter Apr 3 at 9:32
  • $\begingroup$ Okay, I see that, but how does this make all possibilities $9\times9\times3$? $\endgroup$ – Medi Apr 3 at 9:33
  • $\begingroup$ @Medi $9\times 9$ for first, same for second, same for third, hence multiply by $3$. $\endgroup$ – Harry Potter Apr 3 at 9:34
  • $\begingroup$ I like to think of it as choosing $2$ digits that are not $3$s, then placing the $3$. This gives $9 \times 9$ ways to choose the $2$ digits, then we must choose the $3$, which can be placed in $3$ spots (and the other digits are locked in place). With the restriction that $0$ cannot be the first digit, this is basically the same as Light Yagami's answer. $\endgroup$ – Toby Mak Apr 3 at 9:46
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We have $9$ digits other than $0$, but $0$ can't be the first digit. So for both the first and second, we have $8 \cdot 9=72$ possibilities each. For the last one we have $9 \cdot9=81$ possibilities.

Alternatively, there are $9 \times 10 \times 10 = 900$ three-digit numbers. $8 \times 9 \times 9 = 648$ of them have no $3$s as the first digit cannot be $0$ or $3$, and $1$ number has all three $3$s. For the numbers with two $3$s, they must be in the form _ 33, 3 _ 3, 33 _, which makes $8 + 9 + 9 = 26$ possibilities.

Thus there are $900 - 648 - 1 - 26 = 225$ three-digit numbers with exactly one $3$.

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  • $\begingroup$ Thank you for the answer! I appreciate it. $\endgroup$ – Medi Apr 3 at 9:43
  • $\begingroup$ No problem! I'm glad to have helped. $\endgroup$ – Toby Mak Apr 3 at 9:44

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