## shivam_bhalla 3 years ago 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

1. Bdude999

none of these

2. Bdude999

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

3. shivam_bhalla

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

4. shivam_bhalla

5. shivam_bhalla

@precal , any idea??

6. inkyvoyd

Lemme look.

7. inkyvoyd

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

8. shivam_bhalla

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

9. inkyvoyd

So, I can do listing?

10. shivam_bhalla

11. shivam_bhalla

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

12. inkyvoyd

Ok, I think I have an idea.

13. inkyvoyd

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

14. inkyvoyd

4!

15. inkyvoyd

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

16. inkyvoyd

*possibilities

17. inkyvoyd

Actually, find the probability, sorry.

18. shivam_bhalla

I am trying..

19. precal

sorry out of my league

20. shivam_bhalla

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

21. precal

sorry not at the moment

22. shivam_bhalla

Thanks :) Let me keep trying myself :)

23. shivam_bhalla

@amistre64 , any help please ?

24. 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?

25. shivam_bhalla

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

26. shivam_bhalla

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

27. shivam_bhalla

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

28. amistre64

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

29. shivam_bhalla

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

30. 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?

31. amistre64

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

32. shivam_bhalla

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

33. amistre64

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

34. shivam_bhalla

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

35. shivam_bhalla

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??

36. FoolForMath

37. 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?

38. FoolForMath

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

39. shivam_bhalla

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

40. FoolForMath

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

41. shivam_bhalla

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

42. shivam_bhalla

@FoolForMath , Thanks bro, I got it :)

43. FoolForMath

44. shivam_bhalla

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

45. FoolForMath

46. shivam_bhalla

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

47. shivam_bhalla

@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 ??

48. shivam_bhalla

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

49. FoolForMath

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