Number of ways of picking cards

Question

A box contains 10 black cards numbered 1 to 10 and 10 red cards numbered 1 to 10. In how many ways can we choose 10 out of the 20 cards so that there are exactly 3 matches where a match means a red card and a black card with the same number.

My attempt: we can choose any 3 cards from red batch in ${10}\choose{3}$ ways and from the black batch we have to pick the same 3 cards so only one way. Now we have 6 cards and have to choose remaining 4 without having a match, this is what I have not been able to figure out, can someone post the solution.

Answer 1

As you began, first pick which numbers are used for matches in $\binom{10}{3}$ ways. Whichever numbers were picked we will use both the black cards with these numbers and the red cards with these numbers. That accounts for six of our necessary cards.

Now... we can break into cases based on the number of additional red cards which are used. If there are no more red cards used, then there are four additional black cards being used and there is no way for them to match any red cards used. These black cards can be selected in $\binom{7}{4}$ ways. (seven because we have three fewer black cards than normal since those are in use for the matched cards from the first step)

If there was one additional red card used, pick which it is. This can be done in $\binom{7}{1}$ ways. Next, that would mean there are three additional black cards needing to be used, but we don't want any of them to match what we picked for this additional red card. Pick which these are in $\binom{6}{3}$ ways, giving $\binom{7}{1}\times \binom{6}{3}$ outcomes in this case.

Continue in this fashion for two additional, three additional, and four additional red cards being used. Add up the total number of outcomes in this case-work, and then multiply the sum by $\binom{10}{3}$ to get the final total.

$\binom{10}{3}\left(\binom{7}{0}\binom{7}{4}+\binom{7}{1}\binom{6}{3}+\binom{7}{2}\binom{5}{2}+\binom{7}{3}\binom{4}{1}+\binom{7}{4}\binom{3}{0}\right)$

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An alternate way to get the numbers I used in the parentheses.

We begin the same way by picking which numbers are used for matches in $\binom{10}{3}$ ways.

We continue by choosing which four numbers are used for the non-matching cards in $\binom{7}{4}$ ways.

We then choose which color was used for each number in two ways each.

This gives $\binom{10}{3}\binom{7}{4}2^4$