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

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.