In the past I have answered how to distribute 20 different presents to 6 people (of course different people!). Intrinsically two problems are the same. The complexity lies here is the way we draw presents, it is not of necessity to distribute all presents to them.
Ok, now back to this question, just applied the same trick that I have developed in the past, this is an easy problem if one can set up a reccurence relation like the following:
Define $s_n = \{\text{number of way of distribution if exactly \textit{n} people have present(s).}\}$, \[ s_n = ({n+1})^6-\sum_{i=0}^{n-1}\binom{n}{i}s_i,
\] setting $ i=1,2$ and defining $ s_0=1$, one can have that \[
\left\{\begin{array}{l}s_0=1\\s_1 = 2^6 - 1\\s_2 = 3^6 - 2\cdot 2^6 + 1\\s_3 = 4^6 - 3^7 + 3\cdot 2^6 - 1 = 2100\end{array}\right.\]
One may ask what happens when presents are identical, the solution is much trivial and I guess every statistic (especially MAEC ?) student in UST is able to count it.here is the PDF concerning 20 presents case
Click me
No comments:
Post a Comment