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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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FoolForMathBest ResponseYou've already chosen the best response.3
I have already answered you (in that thread)
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.0
Sorry. By mistake I wrote the wrong name :(
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.0
No I am having doubt about which formula to use in which case??
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.3
In this case. The "why" is explained well in the wike page.
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.0
I 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
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.0
Anybody willing to help??
 one year ago

shivam_bhallaBest 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???
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.3
Sorry, right now I can't make things any easier for you.
 one year ago

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