# Math Help - Need help with a few questions.

1. ## Re: Need help with a few questions.

So we choose the 3 suits our hand consists of: 4C3.

Now we choose 5 of the 13 cards of the first suit: 13C5

Now we choose 6 of the 13 cards of the second suit: How do we do this?

Now we choose 2 of the 13 cards of the last suit: How do we do this?

2. ## Re: Need help with a few questions.

Originally Posted by Jame
So we choose the 3 suits our hand consists of: 4C3.

Now we choose 5 of the 13 cards of the first suit: 13C5

Now we choose 6 of the 13 cards of the second suit: How do we do this?

Now we choose 2 of the 13 cards of the last suit: How do we do this?
So 1287, 1716, 78.

Then multiply them all to get a hand?

3. ## Re: Need help with a few questions.

Yes, we multiply

4C3 * 13C5 * 13C6 * 13C2

4. ## Re: Need help with a few questions.

Originally Posted by Jame
Yes, we multiply

4C3 * 13C5 * 13C6 * 13C2
Ok I got 689 049 504, which is the final answer?

5. ## Re: Need help with a few questions.

Yep, that is the answer. You can also leave it 4C3 * 13C5 * 13C6 * 13C2 if you want.

6. ## Re: Need help with a few questions.

Originally Posted by Jame
Yep, that is the answer. You can also leave it 4C3 * 13C5 * 13C6 * 13C2 if you want.
Okay, thanks so much for the help, both you and emakarov!

7. ## Re: Need help with a few questions.

I'm sorry I was actually mistaken in my answer for 2). Since we have different amounts of each suit (5,6,2), we must choose each suit individually. So instead of 4C3, its 4*3*2 (or 4!)

The rest of the answer is the same.

Example: a hand with "5 hearts, 6 spades and 2 diamonds" is different then a hand with "5 diamonds, 6 hearts and 2 spades"
(order of the suits does in fact matter here)

If we had a different breakdown of suits (like 5,5,3). We would have to choose the two suits with the 5 cards (4C2), then choose the suit with 3 cards (2C1)

The only time we would choose the suits all together is if each card had the same breakdown (e.g. (3,3,3))

8. ## Re: Need help with a few questions.

Originally Posted by Jame
I'm sorry I was actually mistaken in my answer for 2). Since we have different amounts of each suit (5,6,2), we must choose each suit individually. So instead of 4C3, its 4*3*2 (or 4!)

The rest of the answer is the same.

Example: a hand with "5 hearts, 6 spades and 2 diamonds" is different then a hand with "5 diamonds, 6 hearts and 2 spades"
(order of the suits does in fact matter here)

If we had a different breakdown of suits (like 5,5,3). We would have to choose the two suits with the 5 cards (4C2), then choose the suit with 3 cards (2C1)

The only time we would choose the suits all together is if each card had the same breakdown (e.g. (3,3,3))
Changed, thanks for letting me know!

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