Combinatorial Analysis (n balls and r urns)

[Win_odd Dhamnekar]

In how many ways can n identical balls be distributed into r urns so that the ith urn contains atleast $\displaystyle m_i$ balls, for each i = 1,. . ., r ? Assume that $\displaystyle n \geq \displaystyle\sum_{i=1}^{r} m_i$

My answer: $\displaystyle \displaystyle\sum_{i=1}^r \binom{n+r -1}{r-1}, n \geq r$

Is this answer correct?

 

[pka]

In how many ways can n identical balls be distributed into r urns so that the ith urn contains atleast $\displaystyle m_i$ balls, for each i = 1,. . ., r ? Assume that $\displaystyle n \geq \displaystyle\sum_{i=1}^{r} m_i$

My answer: $\displaystyle \displaystyle\sum_{i=1}^r \binom{n+r -1}{r-1}, n \geq r$

Is this answer correct?

You want the $i^{th}$ urn to contain at least $m_{i}$ balls.
$M = \sum\limits_{i = 1}^r {{m_i}} \leqslant n$ If we start by puting $m_i$ balls into the $i^{th}$ urn.
Then the remainder of balls can be distributed in $\dbinom{n-M+r-1}{r-1}$ ways.

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