Compute the probability that a hand of $13$ cards contains the ace and king of at least one suit?
Approach 1
In my first approach I took all 8 cards of aces and kings aside and
then made 4 group each containing ace and king of their own suit.
Then for ace and king from 1 suit $\binom{4}1\binom{44}{11}$, for ace and king from 2 suits $\binom{4}2\binom{44}{9}$, for ace and king from 3 suits $\binom{4}3\binom{44}{7}$ and for ace and king from all 4 suits $\binom{4}4\binom{44}{5}$
$$E = \binom{4}1\binom{44}{11} + \binom{4}2\binom{44}{9} + \binom{4}3\binom{44}{7} + \binom{4}4\binom{44}{5}$$
But my first approach is getting me wrong answer.
Approach 2
In this approach I am using inclusion-exclusion identity
$$E = \binom{4}1\binom{50}{11} -\binom{4}2\binom{48}{9} + \binom{4}3\binom{46}{7} -\binom{4}4\binom{44}{5}$$
I am getting right answer by this approach.
But I can't think of how can any cases remained to be not counted in my first approach.
My question what is difference in my both approaches.
