## mathmath333 one year ago A teacher has to distribute 15 pens among 5 of his students such that student A gets at least 3 and at most 6 pens.In how many ways can this be done ?

1. mathmath333

\large \color{black}{\begin{align} & \normalsize \text{A teacher has to distribute 15 pens among 5 of his students such}\hspace{.33em}\\~\\ & \normalsize \text{that student A gets at least 3 and at most 6 pens.In how many }\hspace{.33em}\\~\\ & \normalsize \text{ ways can this be done ?}\hspace{.33em}\\~\\ & a.)\ 495 \hspace{.33em}\\~\\ & b.)\ 77 \hspace{.33em}\\~\\ & c.)\ 417 \hspace{.33em}\\~\\ & d.)\ 435 \hspace{.33em}\\~\\ \end{align}}

2. mathmath333

only student A gets atleast 3 to 6 pens necessarily

3. dan815

lets do 4 cases then Student A 3 4 5 6

4. dan815

when student A gets 3 pens we have 12 pens to distribute among 4 students

5. mathmath333

12C4=495

6. dan815

u cant just do that

7. dan815

why are you doing 12 choose 4

8. mathmath333

cuz i left with 4 students and 12 pens

9. dan815

wait umm from your previous questions, when they ask questions like this does that mean the pens are distinguishable or not?

10. mathmath333

i think pens are similar/identical

11. dan815

okay phew that makes it a lot simpler

12. dan815

so we have 12 pens and 4 students think about it like stars and bars question

13. mathmath333

15C3

14. dan815

|dw:1441579954826:dw|

15. dan815

yes 15 choose 3

16. dan815

now can u do the other cases?

17. mathmath333

14C3 ?

18. dan815

go on

19. mathmath333

15C3+14C3+12C3+11C3 ?

20. dan815

yeah

21. mathmath333

i m dumb

22. mathmath333

u mean it is correct

23. mathmath333

it is 1325 not in options

24. dan815

hm thats weird, i dont see anything wrong with what we did, unless the question should be interpretated differently

25. dan815

@Empty

26. mathmath333

what if the pens are distinct .

27. beginnersmind

You had 15 pens originally but you gave at least 3 to A.

28. dan815

then it iwll be even more

29. beginnersmind

So the highest case is 12C3

30. beginnersmind

and the lowest 9C3

31. anonymous

12C4+11C4+10C4+9C4 is wrong ?

32. dan815

why 12C3?

33. mathmath333

34. mathmath333

yea why 12C3

35. beginnersmind

@Halmos Yes, because you're not choosing people. You are choosing where to put the dividing line between people's stack of pens.

36. beginnersmind

|dw:1441580705091:dw| orginal credit to dan. You started out with 15 pens. But you gave 3 to A. So you have 12 left.

37. dan815

right but after u get 12 pens then u got a total of 15 spots from the 12 pens and 3 divider lines

38. beginnersmind

BTW 12C3 is incorrect too. It should be 11C3. Because there's 11 positions to put your dividing line.

39. beginnersmind

|dw:1441580893991:dw| Or 13. And you can put 2 dividing lines in the same position as well :(

40. beginnersmind

So it's really 3^13 +3^12 +3^11 +3^10

41. dan815

i reallly think its 15 choose 3 for dividing 12 pens between 4ppl if we allow someone to get 0 pens

42. dan815

for example 5 apples between 2 ppl so places = 5 apples + 1 division = 6 places you have 6 places to decide where your division is 6 choose 1 = 6 ways

43. dan815

which is true 0 ,5 1, 4 2, 3 3,2 4,1 5, 0

44. triciaal

student A at least 3 pens 3 or 4 or 5 or 6 next student 12 for 4 students |dw:1441581202316:dw|

45. beginnersmind

So how do you get 15 places from 12 objects?

46. dan815

3 for divisions

47. dan815

what we are about is the order of these 15 things, that determines how its been split up

48. dan815

like if u have 1,1,1,|,| ----- 3 of 1s and 2 bars, the number of distinguishable ways to arrange this will tell you how many ways to distribute 3 identical objects for 3 people

49. dan815

so in this case u have 5 spots for your 2 bars, 5C2

50. beginnersmind

You are confounding choosing positions for your dividers and arranging objects in a certain order.

51. triciaal

|dw:1441581658688:dw|

52. beginnersmind

The way you should think about it is that you have 3 dividers and 13 positions to put the in. More than 1 divider can go in the same position but all the dividers are indistinguishable.

53. dan815

yes that is right

54. dan815

well instead of positions and stuff

55. dan815

you can think about it like 2 distinguishable objets 12 of 1, 3 of the other, and the order they are placed in will determine the unique solutions

56. beginnersmind

That's another way to do it, yes. They should give the same answer. Mine is 3^13, yours is 15!/(12!*3!)

57. beginnersmind

Correction Mine is (3^13)/3!, yours is 15!/(12!*3!)

58. beginnersmind

15!/(12!*3!) = 15C3 Mind Blown

59. triciaal

|dw:1441581910685:dw|

60. beginnersmind

Ok, @dan815, finally figured it out. Your method works, mine doesn't. Sorry, I'm slow tonight.

61. dan815

ah its all good, so why do u think its not in the answers though

62. beginnersmind

Not sure yet.

63. beginnersmind

I'm trying to figure it if there's any double counting going on but I can't see any.

64. beginnersmind

Alright, I give up. Good luck everyone. @mathmath33 you might want to ask your teacher.

65. triciaal

isn't Student A one particular student of the 5? if a student then you have multiples

66. triciaal

or like @beginnersmind calls it "doublecounting"

67. beginnersmind

Well, here's two different related problems. You can try to solve them hoping that one of them gives an answer that's one of the options. Version1: A teacher has to distribute 15 pens among 5 of his students such that there is at least one student who get at least 3 and at most 6 pens.In how many ways can this be done ? Version2: A teacher has to distribute 15 pens among 5 of his students such that there is exactly one student who gets at least 3 and at most 6 pens.In how many ways can this be done ? Version3: A teacher has to distribute 15 pens among 5 of his students such that every student gets at least 3 and at most 6 pens.In how many ways can this be done ? Hint: Version 3 is the easiest :)