I have to calculate number of combination when six cards are chosen from standard 52-card deck, but among these cards it must be at least one card from each suit.
My first thought was (At the beginning I choose 4 important cards and then I choose 2 from the rest):
$${13 \choose 1}{13 \choose 1}{13 \choose 1}{13 \choose 1}{48 \choose 2}$$ but result is more than choose 6 totally random cards, so I can assume it's wrong, but I have no idea why?
Someone suggested me that result is (combinations where 3 cards from one suit plus other combinations):
$${13 \choose 1}{13 \choose 1}{13 \choose 1}{13 \choose 3}{4 \choose 1}+{13 \choose 1}{13 \choose 1}{13 \choose 2}{13 \choose 2}{4 \choose 2}$$
It seems to be also legit for me, then I am so confuse.
To sum up, which result is good?, why?, and why the second one is wrong?
[Math] Combination Choose 6 cards, minimum one from each suit
combinationscombinatorics
Best Answer
In the first answer, you are overcounting because you're imposing an order on the choices you're making. For example, if I choose all four Aces (the first four $\binom{13}{1}$'s), and then choose two Kings - say hearts and spades (the $\binom{48}{2}$), then I get the same collection of cards as if I choose the king of hearts, the king of spades, the ace of clubs, and the ace of diamonds as my "important" cards, and then choose the aces of hearts and spades as the other 2. In fact, that collection of cards gets counted 4 times in your first formula.
The second formula looks good. The first term counts the ways to pick 3 of one suit - $\binom{4}{1}$ chooses which suit gets 3 cards, and the other terms count which cards you get from each suit. The second term is if you have 2 cards from each of 2 suits, and one each from the others. Then $\binom{4}{2}$ counts which suits have 2 cards, and the other terms again pick which cards you get.