5 different games are to be distributes among 4 children randomly. The probabilty that each child get altleast one game is
A)1/4
B)15/64
c) 21/64
d) none of these

- anonymous

- chestercat

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- anonymous

none of these

- anonymous

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

- anonymous

@Bdude999 , probability is always less than or equal to 1 :)

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## More answers

- anonymous

@satellite73 , help please :)

- anonymous

@precal , any idea??

- inkyvoyd

Lemme look.

- inkyvoyd

Sorry, not completely sure how to do this with permutations and combinations. I could do a listing, but that would be quite unefficient.

- anonymous

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

- inkyvoyd

So, I can do listing?

- anonymous

Go ahead

- anonymous

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

- inkyvoyd

Ok, I think I have an idea.

- inkyvoyd

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

- inkyvoyd

4!

- inkyvoyd

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

- inkyvoyd

*possibilities

- inkyvoyd

Actually, find the probability, sorry.

- anonymous

I am trying..

- precal

sorry out of my league

- anonymous

@precal , out of my league too. Do you know who can solve this or who is master of permutations and combinations here??

- precal

sorry not at the moment

- anonymous

Thanks :) Let me keep trying myself :)

- anonymous

@amistre64 , any help please ?

- amistre64

once a child gets a game, are they out of rotation until they all have games? or can it be that 1 child gets all 5 games?

- anonymous

It is like there are 5 different toys and 4 children and we have to find the probability that each child should get atleast one toy

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

Since it is given in the question that we have to find the probability that each child should atleast get one game.

- amistre64

if each child gets a toy and there is one left over; then the prob that each child gets at least 1 toy is: 1 right?

- amistre64

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

- anonymous

Yes you are right. But the question is not that .
It is what is the probability that each boy has atleast one toy.
Which means like
21111 , 12111, 11211,11121,11112 --> These are the favourable cases(accd to question). Now the problem is I need to know total no of cases. to get my probability

- amistre64

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

- anonymous

@amistre64 ,
I have found a pre-written solution to this problem. But I am not able to understand the solution

- anonymous

Here is the solution
n(S) = 4^5
Total ways of distribution so that each child gets atleast one game
\[=4^{5}-4 C _{1} * 3^{5} + 4C_{2} * 2^{5}-4C _{3}\]
= 1024- 4*243 + *32-4 = 240
Reqd probability = 240 / 4^5 = 15/64
Now I don't understand the total ways of distribution in the solution??

- anonymous

@shivam_bhalla Read about Stirling number of second kind

- amistre64

5000
0500
0050
0005 : 4
4100 3200
4010 ...
4001
0410
0401
1400
0041
1040
0140
1004
0104
0014 : 24
0122 0113
0212 ...
0221
1022
1202
1220
2012
2021
2102
2120
2201
2210 : 24
1112
1121
1211
2111 :4
56 altogether? you see any i missed?

- anonymous

The ways of distribution is consistent to dividing r distinct things into n distinct groups \( n! \times S(n,r) \).
If you expand this you will find the closed form which is used in your solution.
REF:http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind#Definition

- anonymous

@FoolForMath , can you suggest a better source because I am totally new to this. Thanks :)
@amistre64 ,Thanks for helping a lot mate :)

- anonymous

Probably this http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html
I haven't (yet) encountered with anything better than these two.

- anonymous

@FoolForMath , thanks a lot mate :) You saved my day :) If I have any doubts regarding this, can I message you ??

- anonymous

@FoolForMath , Thanks bro, I got it :)

- anonymous

Glad to help.

- anonymous

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

- anonymous

Read about "Stars and bars" combinatorics

- anonymous

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

- anonymous

@FoolForMath if this question was dividing 5 common things among four different children, then
the answer would be
\[(5-1)C _{4-1}= 4C _{3} = 4\]
??
And what should be the total no of cases ??

- anonymous

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

- anonymous

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

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