$\newcommand{\si}{\sigma}\newcommand{\Si}{\Sigma}$By rescaling, without loss of generality $\lambda=1$.

The random variable \begin{equation} U_n:=\frac{S-ES}{\binom nk}=\frac1{\binom nk}\sum_{J\in\binom{[n]}k}(Y_J-EY_J) \end{equation} is a U-statistic. Therefore, for each natural $k\ge2$, by Hoeffding's Theorem 7.1, $U_n$ is asymptotically normal (as $n\to\infty$) with (asymptotic) mean $0$ and asymptotic variance $k^2\si_1^2/n$, where \begin{equation} \si_1^2:=Var\,g(X_1),\quad g(X_1):=E(Y_{[k]}|X_1). \end{equation} In our case, \begin{equation} g(x)=1(x\ge t)+1(x<t)(1-q^{k-1}),\quad q:=1-e^{-t}, \end{equation} and hence \begin{equation} \si_1^2=1-q+q \left(1-q^{k-1}\right)^2-\left(1-q^k\right)^2. \end{equation}

It is easy to see that $ES=\binom nk (1-q^k)$.

Thus, $S$ is asymptotically normal with (asymptotic) mean $ES=\binom nk (1-q^k)$ and asymptotic variance $\Si^2:=\binom nk ^2 k^2\si_1^2/n$.

For $n=100$, $k=2$, and $t=1$, we get $S\approx N(ES,\Si^2)$ with $ES\approx2972$ and $\Si:=\sqrt{\Si^2}\approx302$, which is in agreement with your picture.

Explicit bounds on the rate of convergence of $U_n$ to normality (obtained using Stein's method) are available -- see Chen and Shao, Theorem 3.1.

$\newcommand{\si}{\sigma}\newcommand{\Si}{\Sigma}$By rescaling, without loss of generality $\lambda=1$.

The random variable \begin{equation} U_n:=\frac{S-ES}{\binom nk}=\frac1{\binom nk}\sum_{J\in\binom{[n]}k}(Y_J-EY_J) \end{equation} is a U-statistic. Therefore, for each natural $k\ge2$, by Hoeffding's Theorem 7.1, $U_n$ is asymptotically normal (as $n\to\infty$) with (asymptotic) mean $0$ and asymptotic variance $k^2\si_1^2/n$, where \begin{equation} \si_1^2:=Var\,g(X_1),\quad g(X_1):=E(Y_{[k]}|X_1). \end{equation} In our case, \begin{equation} g(x)=1(x\ge t)+1(x<t)(1-q^{k-1}),\quad q:=1-e^{-t}, \end{equation} and hence \begin{equation} \si_1^2=1-q+q \left(1-q^{k-1}\right)^2-\left(1-q^k\right)^2=(1-q)q^{2k-1}. \end{equation}

It is easy to see that $ES=\binom nk (1-q^k)$.

Thus, $S$ is asymptotically normal with (asymptotic) mean $ES=\binom nk (1-q^k)$ and asymptotic variance $\Si^2:=\binom nk ^2 k^2\si_1^2/n$.

For $n=100$, $k=2$, and $t=1$, we get $S\approx N(ES,\Si^2)$ with $ES\approx2972$ and $\Si:=\sqrt{\Si^2}\approx302$, which is in agreement with your picture.

Explicit bounds on the rate of convergence of $U_n$ to normality (obtained using Stein's method) are available -- see Chen and Shao, Theorem 3.1.

$\newcommand{\si}{\sigma}$ The random variable \begin{equation} U_n:=\frac{S-ES}{\binom nk}=\frac1{\binom nk}\sum_{J\in\binom{[n]}k}(Y_J-EY_J) \end{equation} is a U-statistic. Therefore, for each natural $k\ge2$, by Hoeffding's Theorem 7.1, $U_n$ is asymptotically normal (as $n\to\infty$) with (asymptotic) mean $0$ and asymptotic variance $k^2\si_1^2/n$, where \begin{equation} \si_1^2:=Var\,g(X_1),\quad g(X_1):=E(Y_{[k]}|X_1). \end{equation} In our case, \begin{equation} g(x)=1(x\ge t)+1(x<t)(1-q^{k-1}),\quad q:=1-e^{-t}, \end{equation} and hence \begin{equation} \si_1^2=1-q+q \left(1-q^{k-1}\right)^2-\left(1-q^k\right)^2. \end{equation}

It is easy to see that $ES=\binom nk (1-q^k)$.

Thus, $S$ is asymptotically normal with (asymptotic) mean $\binom nk (1-q^k)$ and asymptotic variance $\binom nk ^2 k^2\si_1^2/n$.

Explicit bounds on the rate of convergence of $U_n$ to normality (obtained using Stein's method) are available -- see Chen and Shao, Theorem 3.1.

$\newcommand{\si}{\sigma}\newcommand{\Si}{\Sigma}$By rescaling, without loss of generality $\lambda=1$.

The random variable \begin{equation} U_n:=\frac{S-ES}{\binom nk}=\frac1{\binom nk}\sum_{J\in\binom{[n]}k}(Y_J-EY_J) \end{equation} is a U-statistic. Therefore, for each natural $k\ge2$, by Hoeffding's Theorem 7.1, $U_n$ is asymptotically normal (as $n\to\infty$) with (asymptotic) mean $0$ and asymptotic variance $k^2\si_1^2/n$, where \begin{equation} \si_1^2:=Var\,g(X_1),\quad g(X_1):=E(Y_{[k]}|X_1). \end{equation} In our case, \begin{equation} g(x)=1(x\ge t)+1(x<t)(1-q^{k-1}),\quad q:=1-e^{-t}, \end{equation} and hence \begin{equation} \si_1^2=1-q+q \left(1-q^{k-1}\right)^2-\left(1-q^k\right)^2. \end{equation}

It is easy to see that $ES=\binom nk (1-q^k)$.

Thus, $S$ is asymptotically normal with (asymptotic) mean $ES=\binom nk (1-q^k)$ and asymptotic variance $\Si^2:=\binom nk ^2 k^2\si_1^2/n$.

For $n=100$, $k=2$, and $t=1$, we get $S\approx N(ES,\Si^2)$ with $ES\approx2972$ and $\Si:=\sqrt{\Si^2}\approx302$, which is in agreement with your picture.

Explicit bounds on the rate of convergence of $U_n$ to normality (obtained using Stein's method) are available -- see Chen and Shao, Theorem 3.1.