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Question

# Find the error in the following reasoning: “Consider forming a poker hand with two pairs as a five-step process. Step 1: Choose the denomination of one of the pairs. Step 2: Choose the two cards of that denomination. Step 3: Choose the denomination of the other of the pairs. Step 4: Choose the two cards of that second denomination. Step 5: Choose the fifth card from the remaining denominations. There are$\left( \begin{array} { c } { 13 } \\ { 1 } \end{array} \right)$ways to perform step 1,$\left( \begin{array} { l } { 4 } \\ { 2 } \end{array} \right)$ways to perform step 2,$\left( \begin{array} { c } { 12 } \\ { 1 } \end{array} \right)$ways to perform step 3,$\left( \begin{array} { l } { 4 } \\ { 2 } \end{array} \right)$ways to perform step 4, and$\left( \begin{array} { c } { 44 } \\ { 1 } \end{array} \right)$ways to perform step 5. Therefore, the total number of five-card poker hands with two pairs is$13 \cdot 6 \cdot 12 \cdot 6 \cdot 44 = 247,104$."

Solution

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DEFINITIONS

$\textbf{Multiplication rule: }$If an operation consists of $k$ steps, where the first step can be performed in $n_1$ ways, the second step can be performed in $n_2$ ways, ... and the $k$th step can be performed in $n_k$ ways, then the operation can be performed in $n_1n_2...n_k$ ways.

Definition $\textbf{permutation}$ (order is important):

$P(n,r)=\dfrac{n!}{(n-r)!}$

Definition $\textbf{combination}$ (order is not important):

$C(n,r)=\left(\begin{matrix}n\\ r\end{matrix}\right)=\dfrac{n!}{r!(n-r)!}$

with $n!=n\cdot (n-1)\cdot ...\cdot 2\cdot 1$.

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