Distribution of event times of binomial process $(Y_n)_{n \geqslant 0}$ conditioned on $Y_n = m$ Consider a sequence of independent $0$-or-$1$ outcomes $X_1, \ldots, X_n,\ldots$, such that $\mathbb{P}(X=1)=p$. The discrete time process $\left(Y_n\right)_{n \geqslant 0}$, such that $$Y_n = \sum_{k=1}^n X_k$$
is called binomial process. Let $\tau_i$ denote ordinal numbers where $X_{\tau_i}=1$.  These are event times of the binomial process. Letting $\tau_0=0$, the times between events, $T_i = \tau_{i}-\tau_{i-1}$ are independent and follow Pascal distribution:
$$
    \mathbb{P}\left(T_i = r\right) = p (1-p)^{r-1} [ r \geqslant 1 ]
$$
Clearly $\tau_i < \tau_j$ for all $i<j$.

Q: For a fixed $n > 0$, given $Y_n = m > 0$, I am seeking to determine the distribution of event times $ (\tau_1, \ldots, \tau_m)$.

On other words, for a fixed $0 < m \leqslant n$, I am seeking to determine the distribution of $(T_1, T_1+T_2, \ldots, T_1+\cdots+T_m)$ given that $T_1+\cdots+T_m \leqslant n$ and $T_1+\cdots+T_m +T_{m+1} > n$.
The question is motivated by oft-drawn analogy of binomial process and Poisson process, and the nice corresponding result for the Poisson process, where interarrival times are exponentially distributed, and conditional on there being $m$ events on the interval $(0, \tau)$, arrival times are order statistics of uniform distribution on $(0,\tau)$. I am hoping some similarly neat result could be established for the case of binomial process as well.
 A: Conditionally on $[Y_n=m]$ the set $T=\{\tau_k\mid1\leqslant k\leqslant m\}$ is uniformly distributed on the space $\mathfrak T_m$ of subsets of size $m$ of $\{1,2,\ldots,n\}$.
Proof: for every $t$ in $\mathfrak T_m$, the event $[T=t]$ has absolute probability $p^m(1-p)^{n-m}$. This probability does not depend on $t$ in $\mathfrak T_m$ hence the result follows. In particular,
$$
\mathbb P(T=t\mid Y_n=m)=|\mathfrak T_m|^{-1}={n\choose m}^{-1}.
$$
For example, the number of $t$ in $\mathfrak T_m$ such that $\min t=k$ is
$$
{n-k\choose m-1},
$$
hence
$$
\mathbb P(\tau_1=k\mid Y_n=m)={n-k\choose m-1}\cdot{n\choose m}^{-1}.
$$
A: Although not nearly as elegant as @Did's proof, here is a brute-force one.
Denote the event $\mathcal{E}_{n,m} = \{Y_n=m\}=\{ T_1+\cdots+T_m \leqslant n, T_1+\cdots+T_m + T_{m+1} > n \}$. Then we are after
$$
   \mathbb{P}\left(\tau_1=r_1,\tau_2=r_2,\ldots,\tau_m=r_m \mid \mathcal{E}_{n,m} \right) = \frac{ \mathbb{P}\left(T_1=r_1,T_2=r_2-r_1,\ldots,T_m=r_m-r_{m-1},T_{m+1}>n-r_m\right)}{ \mathbb{P}\left(\tau_m \leqslant n, \tau_m + T_{m+1} > n\right) }  \tag{1}
$$
Using the independence, the numerator reads:
$$\begin{eqnarray}
   \mathbb{P}\left(T_{m+1}>n-r_m\right) \prod_{k=1}^m  \mathbb{P}\left(T_k=r_k - r_{k-1}\right) &=& (1-p)^{n-r_m} \prod_{k=1}^m \left( p (1-p)^{r_k-r_{k-1}-1}\right) \\ &=& p^m (1-p)^{n-m}
\end{eqnarray}
$$
The denominator reads:
$$ \begin{eqnarray}
  \sum_{k=m}^n \mathbb{P}\left(\tau_m=k\right) \mathbb{P}\left(T_{m+1}>n-k\right) &=&  \sum_{k=m}^n \binom{k-1}{m-1} p^m (1-p)^{k-m} (1-p)^{n-k} \\ &=& p^m (1-p)^{n-m} \sum_{k=m}^n \binom{k-1}{m-1}
\end{eqnarray}
$$
Using $\binom{k-1}{m-1} = \binom{k}{m} - \binom{k-1}{m}$ the sum telescopes to give $\binom{n}{m}$.
Substituting expressions for the numerator and the denominator in eq. $(1)$ we recover @Did's result:
$$
   \mathbb{P}\left(\tau_1=r_1,\tau_2=r_2,\ldots,\tau_m=r_m \mid Y_n=m \right) = \frac{p^m (1-p)^{n-m}}{\binom{n}{m} p^m (1-p)^{n-m} } = \frac{1}{\binom{n}{m}}
$$
