We know that the no. of non-negative integral solution to the equation $x_{1} + x_{2} + x_{3} + ... + x_{k} = n$ where $(n≥k)$ and $x_{i} \geq 0, i = \left \{ 1, 2, 3, ..., k \right \}$ is given by $\binom{n \ + \ k \ - \ 1}{k \ - \ 1}$ or $\binom{n \ + \ k \ - \ 1}{n}$.
Here, $m\ (m\geq nk)$ identical balls have to be placed in $n$ distinct bags such that each bag contains at least $k$ balls.
Let, $x_{1}, \ x_{2}, \ x_{3}, \ ..., \ x_{n}$ be the no. of balls in the $n$ distinct bag respectively such that $x_{i} \geq k, i = \left \{ 1, \ 2, \ 3, \ ..., \ n \right \}$. As the total no. of balls is $m$, therefore, $x_{1} + x_{2} + x_{3} + ... + x_{n} = m$.
Now, let $y_{i} = x_{i} - k, \ i = \left \{ 1, 2, 3, ..., n \right \}$. Therefore, if $x_{i} \geq k$, then $y_{i} \geq 0$.
Now, $(x_{1}-k) + (x_{2}-k) + (x_{3}-k) + ... + (x_{n}-k) = m - nk$
$\Rightarrow y_{1} + y_{2} + y_{3} + ... + y_{n} = m - nk$
Now, we have to find all such values of $y_{1}$, $y_{2}$, $y_{3}$,...,$y_{n}$ that satisfy the above equation. In other words, the problem basically reduces to finding the no. of non-negative integral solution to the above equation.
Therefore, the req. no. of ways is $\binom{m \ - \ nk \ + \ n \ - \ 1}{n \ - \ 1} = \binom{m \ - \ nk \ + \ n \ - \ 1}{m \ - \ nk}$.
Therefore, correct option- (B).