Addition and Multiplication in Combinations How can we determine when are we supposed to use addition or multiplication in combinations?
Consider the question:
What is the number of ways of choosing 4 cards from a pack of 52
playing cards? In how many of these
(A) four cards are of the same suit?
(B) four cards belong to four different suits?
In the solution for A, addition between all $4$ combinations have been taken into consideration, whereas for facecards, multiplication has been done. How can we determine which to use when?
Also, in the question:
"two are red cards and two are black cards"
My solution is:
Since there are $26$ red and black cards, to select two red and two black, we will do:
$$52C26 \cdot 26C2+ 52C26 \cdot 26C2$$
But this is wrong why?
 A: 
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.
A: For every selection of $2$ red cards there are $^{26}C_2$ ways to match them with $2$ black cards. As there are also $^{26}C_2$ ways of choosing $2$ red  cards, the number of ways of choosing $2$ red and $2$ black is:
$$n = ^{26}C_2\cdot ^{26}C_2 = 105625$$
For (A) $$n = ^{13}C_4 + ^{13}C_4 + ^{13}C_4 + ^{13}C_4\ \text{or}\  4\cdot ^{13}C_4$$
For (B) $$n = 13\cdot 13\cdot 13\cdot 13 = 13^4$$
