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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

(131)\left( \begin{array} { c } { 13 } \\ { 1 } \end{array} \right)

ways to perform step 1,

(42)\left( \begin{array} { l } { 4 } \\ { 2 } \end{array} \right)

ways to perform step 2,

(121)\left( \begin{array} { c } { 12 } \\ { 1 } \end{array} \right)

ways to perform step 3,

(42)\left( \begin{array} { l } { 4 } \\ { 2 } \end{array} \right)

ways to perform step 4, and

(441)\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

13612644=247,10413 \cdot 6 \cdot 12 \cdot 6 \cdot 44 = 247,104

."

Solution

Verified
Step 1
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DEFINITIONS

Multiplication rule: \textbf{Multiplication rule: }If an operation consists of kk steps, where the first step can be performed in n1n_1 ways, the second step can be performed in n2n_2 ways, ... and the kkth step can be performed in nkn_k ways, then the operation can be performed in n1n2...nkn_1n_2...n_k ways.

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

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

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

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

with n!=n(n1)...21n!=n\cdot (n-1)\cdot ...\cdot 2\cdot 1.

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