I tried the Fundamental Counting principle approach: There are $26$ red cards to chose from for the first card. That leaves $26 \times 25 \times 24 \times 23$ left for the other four cards. So $\frac{26 \times 26 \times 25 \times 24 \times 23}{5!}$ (because we can arrange $5$ cards in $5!$ ways). I get $77740$. Apparently the answer is: $454480$.
I'd appreciate help clearing up my misunderstanding. I'd also like to see the combinations approach. Thanks!
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$\begingroup$Using combinatorics:
We have $\binom{26}{4}\binom{26}{1} + \binom{26}{5} \binom{26}{0}=454480$
The first bit is choosing $4$ black cards and $1$ red card. Then the other scenario is only choosing $5$ red cards! Hope this helps!
$\endgroup$ 1 $\begingroup$You missed two things, I believe:
1) You want $\textit{at most}$ one red card. That means you still have to consider the possibility of all cards being black.
2) When counting the cases in which 4 cards are black and one is red like you did, you don't have $5!$ ways of rearranging, you have less. This is because, as you said, there are $26$ red cards to choose from for the first card. You're never considering the case of choosing a red card $\textit{second}$ or $\textit{third}$, so you shouldn't consider rearrangements in which you swap the red card. Only rearrangements of the $4$ black cards. Does this make sense?
And, in fact, combining these two points:
$$\frac{26\times 25\times 24\times 23\times 22}{5!}+\frac{26\times 26\times 25\times 24\times 23}{4!}=454480$$
I hope it helps.
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