The Principle of inclusion and exclusion to find probability

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A 5-card hard is dealt from a standard deck of 52 cards. Find the Probability of at least 1 heart and 1 spade is among the 5 cards, Using Inclusion and exclusion. So far I have 52C5-2*47C5 (47C5 from the ways to have no spades and no hearts) but from here I am in desperate need of help.

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I see that you are trying to count selections with at least 1 from 13 hearts and at least 1 from 13 spades, by counting the complement for obtaining no hearts or no spades. However, you should be using the Principle of Inclusion and Exclusion. Also your card counts were a little off.

Here's the correction.

$$\def\ch#1#2{{^{#1}\mathrm C_{#2}}}\begin{align}\nu(N_\heartsuit{\geqslant}1\cap N_\spadesuit{\geqslant}1)=& \nu(\Omega)-\nu(N_\heartsuit{=}0\cup N_\spadesuit{=}0)\\[2ex]=&~\nu(\Omega)-\nu(N_\heartsuit{=}0)-\nu(N_\spadesuit{=}0)+\nu(N_{\heartsuit}{=}0\cap N_\spadesuit{=}0) \\[2ex]=&~\ch{52}{5}-2\cdot\ch{(52-13)}{5}+\ch{(52-13\cdot 2)}5\\[2ex]=&~\ch{52}5 -2 \cdot\ch{39}5+\ch{26}5\end{align}$$