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anonymous
 4 years ago
5 pens are to be distributed among 4 children randomly. The probabilty that each child get altleast one pen is ??
@FoolForMath there are two formulas, which are:
\[(k1)C _{n1} \]
where k>Total no of nondistinguishable objects
n>total no of distinguishable receivers
and
\[(n+k1)C _{k} \]
Now can you tell me which formula to use in this case and why?
anonymous
 4 years ago
5 pens are to be distributed among 4 children randomly. The probabilty that each child get altleast one pen is ?? @FoolForMath there are two formulas, which are: \[(k1)C _{n1} \] where k>Total no of nondistinguishable objects n>total no of distinguishable receivers and \[(n+k1)C _{k} \] Now can you tell me which formula to use in this case and why?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have already answered you (in that thread)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry. By mistake I wrote the wrong name :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No I am having doubt about which formula to use in which case??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In this case. The "why" is explained well in the wike page.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I still don't get the difference between theorem 1 and theorem 2 in Stars and bars combinatorics. Can you please make it easier for me to differentiate them

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Anybody willing to help??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@FoolForMath , sorry for troubling you once again, but I am still not getting the difference between theorem 1 and theorem 2 in Stars and bars combinatorics. Can you please make it easier for me to differentiate them ? Please???

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, right now I can't make things any easier for you.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok. No problem . Thanks for prompt reply @FoolForMath I will keep trying it myself :)
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