## shivam_bhalla Group Title 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 2 years ago 2 years ago

1. Bdude999 Group Title

none of these

2. Bdude999 Group Title

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

3. shivam_bhalla Group Title

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

4. shivam_bhalla Group Title

5. shivam_bhalla Group Title

@precal , any idea??

6. inkyvoyd Group Title

Lemme look.

7. inkyvoyd Group Title

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 Group Title

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

9. inkyvoyd Group Title

So, I can do listing?

10. shivam_bhalla Group Title

11. shivam_bhalla Group Title

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

12. inkyvoyd Group Title

Ok, I think I have an idea.

13. inkyvoyd Group Title

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

14. inkyvoyd Group Title

4!

15. inkyvoyd Group Title

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

16. inkyvoyd Group Title

*possibilities

17. inkyvoyd Group Title

Actually, find the probability, sorry.

18. shivam_bhalla Group Title

I am trying..

19. precal Group Title

sorry out of my league

20. shivam_bhalla Group Title

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

21. precal Group Title

sorry not at the moment

22. shivam_bhalla Group Title

Thanks :) Let me keep trying myself :)

23. shivam_bhalla Group Title

@amistre64 , any help please ?

24. amistre64 Group Title

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 Group Title

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 Group Title

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

27. shivam_bhalla Group Title

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

28. amistre64 Group Title

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

29. shivam_bhalla Group Title

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

30. amistre64 Group Title

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 Group Title

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

32. shivam_bhalla Group Title

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 Group Title

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

34. shivam_bhalla Group Title

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

35. shivam_bhalla Group Title

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 Group Title

37. amistre64 Group Title

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 Group Title

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 Group Title

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

40. FoolForMath Group Title

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

41. shivam_bhalla Group Title

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

42. shivam_bhalla Group Title

@FoolForMath , Thanks bro, I got it :)

43. FoolForMath Group Title

44. shivam_bhalla Group Title

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

45. FoolForMath Group Title

46. shivam_bhalla Group Title

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

47. shivam_bhalla Group Title

@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 Group Title

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

49. FoolForMath Group Title

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