Probability that a $5$ cards hand from a standard deck of $52$ contains at least $1$ ace $\frac{C(4, 1)C(51, 4)}{C(52,5)}$? I understand the direct and indirect methods for solving this problem, but I do not understand why $\frac{C(4, 1)C(51, 4)}{C(52,5)}$ is wrong. The answer that $\frac{C(4, 1)C(51, 4)}{C(52,5)}$ results is about 4% bigger than the answer I got with $1 - \frac{C(4, 0)C(48, 5)}{C(52,5)}$ or the direct approach. What are the extra scenarios that $\frac{C(4, 1)C(51, 4)}{C(52,5)}$ incorrectly cover that causes the $4\%$ difference?
 A: You are overcounting.
From what I understand from the formula $\displaystyle \frac{C(4,1)C(51,4)}{C(52,5)}$, you take an ace from any of the four suits, then take any 4 other cards from the remaining.
Let Spades=S, Clubs=C, Hearts=H, Diamond=D, Ace=A, Queen=Q, King=K, Jack=J.
The problem in this is that it counts cases $(1)$ and $(2)$ separately, where

*

*You first select the Ace of Spades. Then you take 4 other cards as H1, SJ, CA, C10.

*You first select th Ace of Clubs. Then you choose H1, SJ, SA, C10.

These two cases (and all other similar cases) are exactly the same but are being counted separately.
A: Since there are
$$\binom{4}{k}\binom{48}{5 - k}$$
ways to select exactly $k$ aces and $5 - k$ non-aces, the correct count of favorable cases should be
$$\binom{4}{1}\binom{48}{4} + \binom{4}{2}\binom{48}{3} + \binom{4}{3}\binom{48}{2} + \binom{4}{4}\binom{48}{1}$$
Your failed attempt counts each hand with more than one ace as many times as an ace appears in the hand, once for each way you could have designated one of the aces in the hand as the ace in that hand.
For instance, by designating a particular ace in the hand as the ace that hand contains, you count the hand $\color{red}{A\heartsuit}, A\spadesuit, \color{red}{7\diamondsuit}, 4\clubsuit, 2\clubsuit$ twice.
\begin{array}{l l}
\text{designated ace} & \text{additional cards}\\ \hline
\color{red}{A\heartsuit} &  A\spadesuit, \color{red}{7\diamondsuit}, 4\clubsuit, 2\clubsuit\\
A\spadesuit & \color{red}{A\heartsuit}, \color{red}{7\diamondsuit}, 4\clubsuit, 2\clubsuit
\end{array}
Similarly, you count the hand $\color{red}{A\heartsuit}, \color{red}{A\diamondsuit}, A\spadesuit, \color{red}{7\diamondsuit}, 4\clubsuit$ three times,
\begin{array}{l l}
\text{designated ace} & \text{additional cards}\\ \hline
\color{red}{A\heartsuit} &  \color{red}{A\diamondsuit}, A\spadesuit, \color{red}{7\diamondsuit}, 4\clubsuit\\
\color{red}{A\diamondsuit} & \color{red}{A\heartsuit}, A\spadesuit, \color{red}{7\diamondsuit}, 4\clubsuit\\
A\spadesuit & \color{red}{A\heartsuit}, \color{red}{A\diamondsuit}, \color{red}{7\diamondsuit}, 4\clubsuit
\end{array}
For the same reason, you count the hand $\color{red}{A\heartsuit}, \color{red}{A\diamondsuit}, A\spadesuit, A\clubsuit, \color{red}{7\diamondsuit}$ four times.
\begin{array}{l l}
\text{designated ace} & \text{additional cards}\\ \hline
\color{red}{A\heartsuit} &  \color{red}{A\diamondsuit}, A\spadesuit, A\clubsuit, \color{red}{7\diamondsuit}\\
\color{red}{A\diamondsuit} & \color{red}{A\heartsuit}, A\spadesuit, A\clubsuit, \color{red}{7\diamondsuit}\\
A\clubsuit & \color{red}{A\heartsuit}, \color{red}{A\diamondsuit}, A\spadesuit, \color{red}{7\diamondsuit}\\
A\spadesuit & \color{red}{A\heartsuit}, \color{red}{A\diamondsuit}, A\clubsuit, \color{red}{7\diamondsuit}
\end{array}
Notice that
$$\binom{4}{1}\binom{48}{4} + \color{red}{2}\binom{4}{2}\binom{48}{3} + \color{red}{3}\binom{4}{3}\binom{48}{2} + \color{red}{4}\binom{4}{4}\binom{48}{1} = \color{red}{\binom{4}{1}\binom{51}{4}}$$
