Video solution:

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Written solution:

The number of distinct hands that can be dealt to the player is equal to: $$\begin{align*} &\binom{52}{10} \\ &= \dfrac{52!}{42!\cdot 10!}\\ &=10\cdot 17\cdot 13\cdot 7\cdot 47\cdot 46\cdot 11\cdot 43\\ &= 158A00A4AA0 \end{align*}$$

Therefore: $$\begin{align*} &17\cdot 13\cdot 7\cdot 47\cdot 46\cdot 11\cdot 43\\ &= 158A00A4AA \end{align*}$$

To find the units digit, we can find the value of this expression mod $10$: $$\begin{align*} A &\equiv 17\cdot 13\cdot 7\cdot 47\cdot \\&46\cdot 11\cdot 43 \bmod{10}\\ &\equiv 7\cdot 3\cdot 7\cdot 7\cdot 6\cdot 1\cdot 3 \bmod{10}\\ &\equiv 1\cdot 9\cdot 6\cdot 3 \bmod{10}\\ &\equiv 9\cdot 8 \bmod{10}\\ &\equiv 2 \bmod{10}\\ \end{align*}$$

Therefore, as $0\le A \le 9,$ and $A\equiv 2\bmod{10},$ it follows that $A=2.$

Thus, A is the correct answer.