# The no. of $4$ Digit no. that contain the Digit $6$ exactly once , si

The no. of $4$ Digit no. that contain the Digit $6$ exactly once , si

My Try:: We Will take the no. from $\left\{0,1,2,3,4,5,6,7,8,9\right\}$

Now for $4$ Digit no.

Case (I) If no. is of the form $\boxed{6}\boxed{+}\boxed{+}\boxed{+}$. Then We can fill these three $+$ sign box by $9\times 8 \times 7 = 504$

Case (II) If no. is of the form $\boxed{+}\boxed{6}\boxed{+}\boxed{+}$. Then We can fill these three $+$ sign box by $8 \times 8 \times 7 = 448$

Case (III) If no. is of the form $\boxed{+}\boxed{+}\boxed{6}\boxed{+}$. Then We can fill these three $+$ sign box by $8 \times 8 \times 7 = 448$

Case (III) If no. is of the form $\boxed{+}\boxed{+}\boxed{+}\boxed{6}$. Then We can fill these three $+$ sign box by $8 \times 8 \times 7 = 448$

So Total $= 504+3 \times 448 = 1848$

But answer Given is $2673$

So Where i have done mistake. plz explain me. Thanks

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The requirement is that 6 appears only once. Other digits can appear more than once.

To wit:

Case (I) If no. is of the form $\boxed{6}\boxed{+}\boxed{+}\boxed{+}$. Then We can fill these three $+$ sign box by $9\times 9 \times 9 = 729$

Case (II) If no. is of the form $\boxed{+}\boxed{6}\boxed{+}\boxed{+}$. Then We can fill these three $+$ sign box by $8 \times 9 \times 9 = 648$

Case (III) If no. is of the form $\boxed{+}\boxed{+}\boxed{6}\boxed{+}$. Then We can fill these three $+$ sign box by $8 \times 9 \times 9 = 648$

Case (III) If no. is of the form $\boxed{+}\boxed{+}\boxed{+}\boxed{6}$. Then We can fill these three $+$ sign box by $8 \times 9 \times 9 = 648$

You have $3*648+729=2673$ possibilities.

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Thanks Russh Got it –  juantheron Jan 31 '13 at 4:03

Remember that the other digits can repeat. So a number like $6999$ is valid.

Your first case should evaluate to $9 \times 9 \times 9 = 729$, and the other three to $8 \times 9 \times 9 = 648$. These sum to $2673$ as you require.

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Thanks Joe Zeng Got it –  juantheron Jan 31 '13 at 4:02