No. of ways of selecting $5$ cards from a deck of of $52$ cards if each selection includes at least one king? Now this question is easy and can be done by considering cases : one king , two kings, three kings or four kings.
That way we have ways
$$= nCr(4,1) \cdot nCr(48,4) + nCr(4,2) \cdot nCr(48,3) + nCr(4,3) \cdot nCr(48,2)+ nCr(4,4) \cdot nCr(48,1)$$
However, 
what is conceptually wrong in the following approach:
"Say, we select one king out of $4$ in $nCr(4, 1)$ way and then mix up the remaining cards and the remaining $3$ kings and select four out of them in $nCr(51,4)$ ways.
Making total no. of ways $=nCr(4,1) \cdot nCr(51,4)$
Essentially we are fixing one of the king in our selection and selecting $4$ out of $51$ cards(as a whole).
 A: You are counting hands with more than one king multiple times.
Suppose you have $\color{red}{K\heartsuit}, \color{red}{K\diamondsuit}, 10\clubsuit, 7\spadesuit, \color{red}{5\diamondsuit}$.  You count this hand twice, once when you designate the $\color{red}{K\heartsuit}$ as the designated king and once when you designate the $\color{red}{K\diamondsuit}$ as the designated king.
$$\color{red}{K\heartsuit} \qquad \color{red}{K\diamondsuit}, 10\clubsuit, 7\spadesuit, \color{red}{5\diamondsuit}$$
$$\color{red}{K\diamondsuit} \qquad \color{red}{K\heartsuit}, 10\clubsuit, 7\spadesuit, \color{red}{5\diamondsuit}$$
More generally, you count hands with two kings twice, three kings three times, and four kings four times.  Observe that 
$$\binom{4}{1}\binom{48}{4} + \binom{2}{1}\binom{4}{2}\binom{48}{3} + \binom{3}{1}\binom{4}{3}\binom{48}{2} + \binom{4}{1}\binom{4}{4}\binom{48}{1} = \binom{4}{1}\binom{51}{4}$$
A: When you see the words "at least", it is a safe bet that the easiest way to solve the problem will be to find the complementary set. In this case, this means to find the number of hands with no king, and subtract it from the total number of hands.
A: The method above counts some hands twice. This usually indicates a need for Inclusion-Exclusion. Let $S_\spadesuit,S_\heartsuit,S_\diamondsuit,S_\clubsuit$ be the collection of hands with the corresponding kings.
$$
N(j)=\sum_{|A|=j}\left|\,\bigcap_{i\in A} S_i\,\right|
$$
Then
$$
N(j)=\overbrace{\ \ \ \binom{4}{j}\ \ \ }^{\substack{\text{number of ways}\\\text{to choose $j$ kings}}}\ \overbrace{\binom{52-j}{5-j}}^{\substack{\text{after choosing $j$}\\\text{kings, the number}\\\text{of ways to choose}\\\text{the rest of the cards}}}
$$
Inclusion-Exclusion says that the number of hands with at least one king is
$$
\begin{align}
\sum_{j=1}^4(-1)^{j-1}N(j)
&=\binom{4}{1}\binom{51}{4}-\binom{4}{2}\binom{50}{3}+\binom{4}{3}\binom{49}{2}-\binom{4}{4}\binom{48}{1}\\
&=886656
\end{align}
$$

Another method is to count the hands with $k$ kings separately:
$$
\overbrace{\binom{4}{1}\binom{48}{4}}^\text{$1$ king}+\overbrace{\binom{4}{2}\binom{48}{3}}^\text{$2$ kings}+\overbrace{\binom{4}{3}\binom{48}{2}}^\text{$3$ kings}+\overbrace{\binom{4}{4}\binom{48}{1}}^\text{$4$ kings}=886656
$$

Yet another method is to subtract the number of hands with no kings from the total number of hands:
$$
\binom{52}{5}-\binom{48}{5}=886656
$$
