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I purchased $5$ decks of cards from the store and shuffled them all together. Now I have $260$ mixed cards ($5$ decks $\times$ $52$ cards).

What is the minimum number of cards I need to take out from the $260$ cards to get a full house?

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    Well, there are $13$ ranks in total so take two from each and you have a $26$ card bad hand...adding one more guarantees a full house, so $27$. – 2017-02-12
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    I'm confident that the minimum is $5$ cards for a full house. However if you want the minimum number to *guarantee* a full house, I defer to @lulu – 2017-02-12
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    @Joffan Exactly. though I think there might be a gap in what I wrote. For example, you might draw $20$ aces....in which case you could have a bad hand with $20$ aces, and one from each other rank. Thus a $32$ card bad hand. – 2017-02-12

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The solution I sketched in the comments is too hasty. Assuming we do not count five of a kind as a full house, then the worst hand you can draw would have $20$ of one rank, and one each of all the other ranks. Thus $20+12=32$ cards in total. Therefore the minimum required number is actually $\fbox {33}$.

To check carefully that $33$ works, suppose we had a "bad" hand with $33$ cards.

First note that you must have at least three cards of some given rank. This is because $2\times 13=26<33$. Say we have $N≥3$ of that given rank. Of course we can't have more than $1$ card in each of the following ranks. Suppose that we have exactly one of each of $J$ suits, $J≤12$. Thus we have $N+J$ cards. It follows that $N+J=33\implies N≥21$. But we can't have $21$ cards of a given rank, so we have a contradiction.