smc

We want the number of Successful Outcomes over the number of Total Outcomes. We want to calculate the total outcomes first. Since we have $12$ coins and we need to choose $6$, we have $\left(\genfrac{}{}{0ex}{}{12}{6}\right)$ = $924$ Total outcomes. For our successful outcomes, we can have $(1)1$ penny and $5$ dimes, $2$ nickels and $4$ dimes, $1$ nickel and $5$ dimes, or $6$ dimes. For the case of $1$ penny and $5$ dimes, there are $\left(\genfrac{}{}{0ex}{}{6}{5}\right)$ ways to choose the dimes and $2$ ways to choose the pennies. That is $6\cdot 2=12$ successful outcomes. For the case of $2$ nickels and $4$ dimes, we have $\left(\genfrac{}{}{0ex}{}{6}{4}\right)$ ways to choose the dimes and $\left(\genfrac{}{}{0ex}{}{4}{2}\right)$ ways to choose the nickels. We have $15\cdot 6$ = $90$ successful outcomes. For the case of $1$ nickel and $5$ dimes, we have $\left(\genfrac{}{}{0ex}{}{4}{1}\right)\cdot \left(\genfrac{}{}{0ex}{}{6}{5}\right)=24$. Lastly, we have $6$ dimes and $0$ nickels, and $0$ pennies, so we only have one case. Therefore, we have $\frac{12+90+24+1}{924}=\frac{127}{924}$ = $\boxed{\frac{127}{924}}$