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none of these

the chance a child gets a game is more than one, all the fractions are less than one

@satellite73 , help please :)

Lemme look.

no problem. I have also faced difficulties in this problem :)

So, I can do listing?

Go ahead

But I warn that it would be too hectic. Permutation and Combination would be the right way

Ok, I think I have an idea.

First find the probability that the 4 games would be distributed evenly.

4!

Now, find the probability that ou of the 4 games, one got repeated.

*possibilities

Actually, find the probability, sorry.

I am trying..

sorry out of my league

sorry not at the moment

Thanks :) Let me keep trying myself :)

@amistre64 , any help please ?

or can it be that 1 child gets all 5 games --->not possible

It can be that 1 child gets 2 toys and rest of them get one each

if the toys are handed out at random, then i dont see why 1 child couldnt get all the toys

otherwise there has to be the possibility that one child gets all the toys and such

well,
1112
1121
1211
2111 is your favoured cases
and i we bruting out the total cases

@shivam_bhalla Read about Stirling number of second kind

@FoolForMath , Thanks bro, I got it :)

Glad to help.

If this question was diving 5 common things among four different children, then??

Read about "Stars and bars" combinatorics

Will read about Stars and bars" combinatorics and report back here :)

@FoolForMath Or the answer is
\[(5+4-1) C _{4-1}= 8 C _{3}=56\]

with the at-least one constraint it will be the first one.