How many 13-card hands have at least one Jack, King, Queen, or Ace?

Question

So with this question, I came to this math:

I have a J, Q, K, and A in four suits, and after having one of those face cards in a hand, now we are left to choose 12 more cards. so then I figure we get 4$4 \choose 1$ $48 \choose 12$.

However, I believe I am overcounting, I just don't know how. Can someone help me?

Answer 1

Yes, you are.   Either use the principle of inclusion and exclusion, or alternatively look to the complement event: having not any of those sixteen cards. (Hint: The later is easier.)

$$\tbinom {16} 1\tbinom {36} {12} -\tbinom {16} 2\tbinom {36} {11}+\tbinom {16} 3\tbinom {36} {10}-\tbinom{16}{4}\tbinom{36}{9} + \tbinom{16}{5}\tbinom{36}{8} - \tbinom{16}{6}\tbinom{36}{7}+\tbinom{16}{7}\tbinom{36}{6}-\tbinom{16}{8}\tbinom{36}{5}+\tbinom{16}{9}\tbinom{36}{4}-\tbinom{16}{10}\tbinom{36}{3}+\tbinom{16}{11}\tbinom{36}{2}-\tbinom{16}{12}\tbinom{36}{1}+\tbinom{16}{13} \\[3ex] \;=\;\binom{52}{13} - \binom{36}{13}$$