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

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?

• 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. – Shubham Johri Apr 3 at 9:27
• Are you open to alternative ways to approach this question, or do you just want to know why your steps were wrong? – Toby Mak Apr 3 at 9:28
• @TobyMak, I am open to alternative ways to approach this question! But of course I want to know why my steps are wrong. – Medi Apr 3 at 9:29
• Never mind, Light Yagami beat me to it. – Toby Mak Apr 3 at 9:30
• You have probably ignored that digits other than $3$ can be repeated – Shubham Johri Apr 3 at 9:30

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 $$243-18=225$$.

• Aren't all possibilities to choose 3 from 10 digits $V_{10}^3=10\times9\times8$? – Medi Apr 3 at 9:31
• @Medi 3 cannot be used again and rest all digits can be repeated. – Harry Potter Apr 3 at 9:32
• Okay, I see that, but how does this make all possibilities $9\times9\times3$? – Medi Apr 3 at 9:33
• @Medi $9\times 9$ for first, same for second, same for third, hence multiply by $3$. – Harry Potter Apr 3 at 9:34
• 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. – Toby Mak Apr 3 at 9:46

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 $$3$$.

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