[Math] In how many ways could you choose 5 cards with 2 hearts or 3 spades

combinatorics

For all I know that there are $52$ cards $13$ of every type. So we can choose $2$ hearts in $\binom{13}{2}$ ways and $3$ spades in $\binom{13}{3}$ ways, so we can choose both in $\binom{13}{2}\cdot\binom{13}{3}$, but what about either this one or this one but not both (OR)
I think since we have in the hearts case one way to choose the first two spots and $\binom{39}{3}$ ways for the rest, in the second case we one way for the last three spots and $\binom{39}{2}$ for the rest. Therefor we have $\binom{39}{2} + \binom{39}{3} – \binom{13}{2} \cdot \binom{13}{3}= 741 + 9139 – 22308$ which is negative so what did I do wrong??

Best Answer

Hints:

  • Choosing $5$ cards such that there are exactly $2$ hearts can be done on $\binom{13}2\binom{39}3$ ways.
  • $\mathsf P(A\cup B)=\mathsf P(A)+\mathsf P(B)-\mathsf P(A\cap B)$ (applying on OR)
  • $\mathsf P(A\Delta B)=\mathsf P(A)+\mathsf P(B)-2\mathsf P(A\cap B)$ (applying on XOR)
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