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

[imath][/imath]