Probability - throwing a dice ten times 
A fair dice was thrown 10 times. What is the probability that at least one of the numbers 1-6 appeared exactly 5 times?

Can someone help me?
 A: The probability that a given number is thrown exactly $5$ times is
$$
\overbrace{\ \ \binom{10}{5}\ \ }^{\substack{\text{the number}\\\text{of ways to}\\\text{arrange the}\\\text{two groups}}}\overbrace{\ \ \left(\frac16\right)^5\ }^{\substack{\text{probability}\\\text{of choosing}\\\text{five of the}\\\text{given number}}}\overbrace{\ \ \left(\frac56\right)^5\ }^{\substack{\text{probability}\\\text{of choosing}\\\text{five of the}\\\text{other numbers}}}
$$
There are $\binom{6}{1}$ possibilities for the given number.
The probability that two given numbers are thrown exactly $5$ times is
$$
\overbrace{\ \ \binom{10}{5}\ \ }^{\substack{\text{the number}\\\text{of ways to}\\\text{arrange the}\\\text{two groups}}}\overbrace{\ \ \left(\frac16\right)^5\ }^{\substack{\text{probability}\\\text{of choosing}\\\text{five of the}\\\text{first number}}}\overbrace{\ \ \left(\frac16\right)^5\ }^{\substack{\text{probability}\\\text{of choosing}\\\text{five of the}\\\text{other number}}}
$$
There are $\binom{6}{2}$ possibilities for the given numbers.
Inclusion-Exclusion says that the probability that at least one number is thrown $5$ times is
$$
\binom{6}{1}\binom{10}{5}\left(\frac16\right)^5\left(\frac56\right)^5-\binom{6}{2}\binom{10}{5}\left(\frac16\right)^{10}=\frac{43715}{559872}
$$
