Question

Hey folks, a few more combinatorics questions:

1. From a deck of 52 playing cards, how many different 4-card hands can be formed with at most 1 diamond?

2. To play Lotto 649 a person chooses 6 numbers out of a possible 49. The Lottery Corporation then randomly chooses 6 numbers as well. How many possible six number sets can be chosen by the Lottery Corporation that match exactly 4 of the numbers that you have?

Answer 1

For the first question I thought it would be (13C1*39C4)+(13C0*39C5), but apparently not. For the second I'm really not sure where to start - I tried dividing 49C6 by 49C5 but that's not getting me anywhere.

Answer 2

For the first problem, you almost got it. What you're doing wrong, is choosing 5 cards instead of 4. You should have\[13C1\cdot39C3+13C0\cdot39C4\]

Answer 3

http://mkwan08.files.wordpress.com/2011/09/combinations.pdf

Answer 4

page 2 and 3 of this pdf shows and explains these 2 problems.

Answer 5

The second part is a bit harder. Give me a minute or two.

Answer 6

For the second problem have we already chosen six numbers?

Answer 7

And do each of the numbers have to be distinct?

Answer 8

I don't have any more info than what's given. It states "a person choses 6 numbers out of a possible 49" So I assume that's just 6C49.

Thanks by the way, not sure how I missed that for the first question..

Answer 9

In that case, I'll assume that the numbers are distinct, order doesn't matter, and you haven't already chosen your 6 numbers.

Answer 10

First, \(49C6\). That much should be obvious. Now, 4 of these must match what the lottery company chooses. That means we multiply by \(6C2\).

Answer 11

Next, the lottery company still needs to choose 2 more numbers, but they can't match the other two numbers that you have, and they don't repeat. That means you multiply by \[43C2\]Since 49-6=43.

Answer 12

This leaves you with a final answer of \[\binom{49}{4}\binom{43}{2}\binom{6}4\]I hope that's correct :/

Answer 13

I feel like this isn't correct. Not sure why though...

Answer 14

Hmm, unfortunately not, your answer gives 2869860420 but the answer key says it's 13545.

Answer 15

I see why it's not right. We're counting the sets that the lottery company chooses. So this method is overcounting by a lot.

For example, you pick {1,2,3,4,5,6},{1,2,3,4,9,10},{1,2,5,6,7,8},... and lotto picks {1,2,3,4,7,8}. So we're overcounting by a lot.

Answer 16

I think i just figured it out - it's just the four numbers the lottery needs to choose multiplied by the possibilities for the two other numbers:

\[\left(\begin{matrix}43 \\ 2\end{matrix}\right)\left(\begin{matrix}6 \\ 4\end{matrix}\right)=13545\]

I think I've got it now, didn't realize I had to consider those other two numbers as well. Thanks!

Answer 17

Of course. That looks right.