Concept Used:
Number of non-negative integral solution to the equation
$x_{1} + x_{2} + x_{3} + ... + x_{k} = n$
where, $(n≥k)$,
$x_{i} \geq 0,$ and
$i = \left \{ 1, 2, 3, ..., k \right \}$
$=\binom{n \ + \ k \ - \ 1}{k \ - \ 1}$ or $\binom{n \ + \ k \ - \ 1}{n}$.
Problem:
$m\ (m\geq nk)$ identical balls have to be placed in $n$ distinct bags such that each bag contains at least $k$ balls.
Solution:
Let $Bag_i$ contains $x_i$ number of balls where $i = \left \{ 1, \ 2, \ 3, \ ..., \ n \right \}$
Then sum of all the balls in each bag = $m$ where $(m\geq nk)$
$\implies$ $x_{1} + x_{2} + x_{3} + ... + x_{n} = m$.
$\because$ Each bag should contain atleast $k$ number of balls so we can write each of the $x_i$'s as sum of $y_i + k$
$\implies$ $(y_{1}+k) + (y_{2}+k) + (y_{3}+k) + ... + (y_{n}+k) = m$.
$\implies$ $y_{1} + y_{2} + y_{3} + ... + y_{n} = m - nk$.
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 number of non-negative integral solution to the above equation.
$\therefore$ The required number of ways is $\binom{m \ - \ nk \ + \ n \ - \ 1}{n \ - \ 1} = \binom{m \ - \ nk \ + \ n \ - \ 1}{m \ - \ nk}$.
Therefore, correct option- (B).
