What is the number of ways of choosing four cards of the same suit when four cards are selected from a standard deck of $52$ playing cards?
Method 1: To be successful, we must choose four of the thirteen clubs or four of the thirteen diamonds or four of the thirteen hearts or four of the thirteen spades, which can be done in
$$\binom{13}{4} + \binom{13}{4} + \binom{13}{4} + \binom{13}{4} = 4\binom{13}{4}$$
ways.
Why do we add?
The Addition Principle states that if there are $m$ ways of performing one task and $n$ ways of performing another task that cannot be done at the same time, then the number of ways of performing one of the tasks is $m + n$.
In this case, we may choose four clubs or four diamonds or four hearts or four spades. Since these tasks are mutually exclusive (cannot be performed at the same time), we add the number of ways each task can be performed.
The word or is an indication that you need to add.
Method 2: To be successful, we must choose one of the four suits and four of the thirteen cards of that suit in
$$\binom{4}{1}\binom{13}{4}$$
ways.
Why do we multiply?
The Multiplication Principle states that if one task can be performed in $m$ ways and a second task can be performed independently of the first in $n$ ways, then there are $mn$ ways of performing both tasks.
In this case, we choose a suit and then choose four cards of that suit. Since the number of ways we can select four cards of the chosen suit is independent of the choice of suit, we multiply the number of ways each task can be performed.
The word and is an indication that you have to multiply.
What is the number of ways of choosing four cards of different suits when four cards are selected from a standard deck of $52$ cards?
We must choose one of the thirteen clubs and one of the thirteen diamonds and one of the thirteen hearts and one of the thirteen spades, which can be done in
$$\binom{13}{1}\binom{13}{1}\binom{13}{1}\binom{13}{1}$$
ways.
Notice that each of the four tasks is performed independently of the others, so we multiply the number of ways each task can be performed.
What is the number of ways of choosing two red and two black cards when four cards are selected from a standard deck of $52$ playing cards?
We must select two of the twenty-six red cards and two of the twenty-six black cards, which can be done in
$$\binom{26}{2}\binom{26}{2}$$
ways.
Notice that each task is performed independently of the other, so we multiply the number of ways each task can be performed.
Why is your answer wrong?
You first selected $26$ of the $52$ cards in the deck, then multiplied by the number of ways to select two of those $26$ cards of one of the colors. However, $\binom{52}{26}$ is the number of ways of selecting any $26$ cards in the deck (which do not all need to be the same color). Instead, you should have chosen all $26$ cards of that color and then chosen two cards of that color for each color, which can be done in
$$\binom{26}{26}\binom{26}{2} = \binom{26}{2}$$
ways.
Moreover, you have to this for the red cards and the black cards, so you have to multiply the number of ways of doing this for each color.
Best Answer
I don't believe you'll find a general algorithm for solving such problems, you tackle each on its own.
For each of the 13 different ranks you have 4 available cards.
How many ways are there to draw 3 cards of the same rank? Answer: 13 times the number of ways to draw 3 cards of any specific rank. Which means, $13 \times \binom{4}{3} = 13 \times 4 = 52$.
The question about different ranks needs a different solution. You need to choose the 3 different ranks first, and then select one of the 4 available cards for each selected rank. For the first choice we have $\binom{13}{3}$ ways, and for each of the three ranks you have 4 possible cards. In total, $\binom{13}{3} \times 4 \times 4 \times 4 = \frac{13!}{3!\times10!}\times 4^3=18\,304$.
It might help if you compare the strategies we used:
Now I'll answer your comment's question about the ways to have 2 cards of the same rank and 1 of a different one.
What you suggest, $\binom{13}{2} * 4 * 4 * \binom{13}{1} * 4$, does not really count what you want.
With $\binom{13}{2}$ you choose 2 different ranks, then $4 * 4$ gives you a card from each one, and then $\binom{13}{1} * 4$ chooses a single card at random from the entire 52-card deck, without considering that the previously drawn cards are out of it.
I.e. your expression solves "How many ways are there to draw 2 differently ranked cards from a single deck and a random card from a different deck".
You want to solve "How many ways are there to draw 2 cards of the same rank and a card of a different rank".
For this, you need 2 ranks - one for the two equally ranked cards, and one for the third. You may think $\binom{13}{2}$ works, but this gives you two "unordered"1 ranks, which means that "a King and a Queen" and "a Queen and a King" count as the same draw. However, we assign meaning to the first and to the second rank - one will have 2 cards, and the other only one. So the actual number of ways to choose the 2 ranks here is twice as big as $\binom{13}{2}$.
Finally, choose the two same-rank cards $\binom{4}{2}$ and the third one $\binom{4}{1}$. In total you get $2 \times \binom{13}{2} \times \binom{4}{2} \times \binom{4}{1}$, or $3744$.
1 If someone finds a better term, please share it.