\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
There are \(\binom {2n}{2k}\) ways to select the socks in total.
All \(2k\) socks must be from different pairs of socks, so we have to select \(2k\) pairs of socks from the \(n\) pairs available, giving us \(\binom {n}{2k}\) options.
Out of those \(2k\) pairs, one of the two is selected, which gives \(2^{2k}\).
Therefore, the probability is given by \[ \Prob = \frac {\binom {n}{2k} \cdot 2^{2k}}{\binom {2n}{2k}}. \]
There are \(r\) pairs of socks, and \(2k - 2r = 2 (k - r)\) socks that do not form any pairs (single).
This gives us \(\binom {n}{r}\) to select the \(r\) pairs of socks, \(\binom {n - r}{2(k - r)}\) to select the \(2(k - r)\) pairs from the remaining \(n - r\) pairs. Finally, there is a factor of \(2^{2 (k - r)}\) ways to select one sock out of the \(n - r\) pair.
Hence, \[ \Prob (X_{n, k} = r) = \frac {\binom {n}{r} \binom {n - r}{2 (k - r)} 2^{2 (k - r)}}{\binom {2n}{2k}} \] as desired, for \(0 \leq r \leq k\).
By expanding out the binomial coefficients, we have \begin {align*} \Prob (X_{n, k} = r) & = \frac {\frac {n!}{(n - r)! r!} \cdot \frac {(n - r)!}{(2(k - r))! ((n - r) - 2(k - r))!}}{\frac {(2n)!}{(2k)! (2(n - k))!}} \cdot 2^{2 (k - r)} \\ & = \frac {n! (2k)! (2(n - k))!}{(2n)! r! (2(k - r))! (n + r - 2k)!} \cdot 2^{2 (k - r)}, \end {align*}
and hence \begin {align*} & \phantom {=} \Prob (X_{n - 1, k - 1} = r - 1) \\ & = \frac {(n - 1)! (2 (k - 1))! (2((n - 1) - (k - 1)))!}{(2(n - 1))! (r - 1)! (2((k - 1) - (r - 1)))! ((n - 1) + (r - 1) - 2(k - 1))!} \cdot 2^{2 ((k - 1) - (r - 1))} \\ & = \frac {(n - 1)! (2k - 2)! (2(n - k))!}{(2n - 2)! (r - 1)! (2(k - r))! (n + r - 2k)!} \cdot 2^{2 (k - r)}. \end {align*}
To show that \[ r \cdot \Prob (X_{n, k} = r) = \frac {k (2k - 1)}{2n - 1} \cdot \Prob (X_{n - 1, k - 1} = r - 1), \] it is equivalent to showing that \begin {align*} r \cdot \frac {n! (2k)!}{(2n)! r! } & = \frac {k (2k - 1)}{2n - 1} \cdot \frac {(n - 1)! (2k - 2)! }{(2n - 2)! (r - 1)!} \\ r \cdot \frac {n (2k) (2k - 1)}{(2n) (2n - 1) r } & = \frac {k (2k - 1)}{2n - 1} \\ \frac {n (2k)}{2n} & = k \\ 2nk & = 2nk \end {align*}
which is true.
Therefore, we have \[ r \cdot \Prob (X_{n, k} = r) = \frac {k (2k - 1)}{2n - 1} \cdot \Prob (X_{n - 1, k - 1} = r - 1) \] as desired.
Therefore, the expectation can be simplified as \begin {align*} \Expt (X_{n, k}) & = \sum _{r = 0}^{k} r \Prob (X_{n, k} = r) \\ & = \sum _{r = 1}^{k} r \Prob (X_{n, k} = r) \\ & = \sum _{r = 1}^{k} \frac {k (2k - 1)}{2n - 1} \Prob (X_{n - 1, k - 1} = r - 1) \\ & = \frac {k (2k - 1)}{2n - 1} \sum _{r = 1}^{k} \Prob (X_{n - 1, k - 1} = r - 1) \\ & = \frac {k (2k - 1)}{2n - 1} \sum _{r = 0}^{k - 1} \Prob (X_{n - 1, k - 1} = r - 1) \\ & = \frac {k (2k - 1)}{2n - 1} \cdot 1 \\ & = \frac {k (2k - 1)}{2n - 1} \end {align*}
since \(0 \leq X_{n, k} \leq k\), \(0 \leq X_{n - 1, k - 1} \leq k - 1\) and that they can only take integer values.