Genesis Gibbs

2022-09-06

I tried the Fundamental Counting principle approach: There are 26 black 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!

I'd appreciate help clearing up my misunderstanding. I'd also like to see the combinations approach. Thanks!

Kyan Mcdonald

Beginner2022-09-07Added 6 answers

You missed two things, I believe:

1) You want at most one black 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 black like you did, you don't have 5! ways of rearranging, you have less. This is because, as you said, there are 26 black cards to choose from for the first card. You're never considering the case of choosing a black card second or third, so you shouldn't consider rearrangements in which you swap the black card. Only rearrangements of the 4 black cards. Does this make sense?

And, in fact, combining these two points:

$\frac{26\times 25\times 24\times 23\times 22}{5!}+\frac{26\times 26\times 25\times 24\times 23}{4!}=454480$

1) You want at most one black 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 black like you did, you don't have 5! ways of rearranging, you have less. This is because, as you said, there are 26 black cards to choose from for the first card. You're never considering the case of choosing a black card second or third, so you shouldn't consider rearrangements in which you swap the black card. Only rearrangements of the 4 black cards. Does this make sense?

And, in fact, combining these two points:

$\frac{26\times 25\times 24\times 23\times 22}{5!}+\frac{26\times 26\times 25\times 24\times 23}{4!}=454480$

clovnerie0q

Beginner2022-09-08Added 1 answers

Using combinatorics:

We have $(}\genfrac{}{}{0ex}{}{26}{4}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{26}{1}{\textstyle )}+{\textstyle (}\genfrac{}{}{0ex}{}{26}{5}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{26}{0}{\textstyle )}=454480$

The first bit is choosing 4 black cards and 1 black card. Then the other scenario is only choosing 5 black cards! Hope this helps!

We have $(}\genfrac{}{}{0ex}{}{26}{4}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{26}{1}{\textstyle )}+{\textstyle (}\genfrac{}{}{0ex}{}{26}{5}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{26}{0}{\textstyle )}=454480$

The first bit is choosing 4 black cards and 1 black card. Then the other scenario is only choosing 5 black cards! Hope this helps!

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