probability that exactly three of them shows the same face and remaining three shows different faces Question

$6$ dice are thrown simultaneously ,find the probability that exactly three of them shows the same face and remaining three shows different faces.

My Approach
Total outcome=$$6^{6}$$
Number of ways to select an outcome which will be same$$=\binom{6}{1}$$
so selected outcome will be $$=\frac{ \binom{6}{1}\times 1 \times 1 \times 1 \times 5 \times4 \times 3}{6^{6}}=\frac{6\times5 \times4\times 3 }{6^6}$$
Am i correct?

One may see it as follows:


*

*There are 6 options for the equal faces

*$\binom{6}{3}$ choices, which 3 dice show the same face

*$5\cdot 4 \cdot 3$ choices for the mutually different faces of the other three

*all together there are $6^6$ different throws
$$\frac{6\cdot \binom{6}{3}\cdot 5\cdot 4\cdot 3}{6^6}= \frac{25}{162}$$

