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 ?

- mathmath333

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- mathmath333

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

- mathmath333

only student A gets atleast 3 to 6 pens necessarily

- dan815

lets do 4 cases then
Student A
3
4
5
6

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## More answers

- dan815

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

- mathmath333

12C4=495

- dan815

u cant just do that

- dan815

why are you doing 12 choose 4

- mathmath333

cuz i left with 4 students and 12 pens

- dan815

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

- mathmath333

i think pens are similar/identical

- dan815

okay phew that makes it a lot simpler

- dan815

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

- mathmath333

15C3

- dan815

|dw:1441579954826:dw|

- dan815

yes 15 choose 3

- dan815

now can u do the other cases?

- mathmath333

14C3 ?

- dan815

go on

- mathmath333

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

- dan815

yeah

- mathmath333

i m dumb

- mathmath333

u mean it is correct

- mathmath333

it is 1325 not in options

- dan815

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

- dan815

@Empty

- mathmath333

what if the pens are distinct .

- beginnersmind

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

- dan815

then it iwll be even more

- beginnersmind

So the highest case is 12C3

- beginnersmind

and the lowest 9C3

- anonymous

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

- dan815

why 12C3?

- mathmath333

answer given is d.) 435

- mathmath333

yea why 12C3

- beginnersmind

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

- 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.

- dan815

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

- beginnersmind

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

- beginnersmind

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

- beginnersmind

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

- dan815

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

- 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

- dan815

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

- triciaal

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

- beginnersmind

So how do you get 15 places from 12 objects?

- dan815

3 for divisions

- dan815

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

- 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

- dan815

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

- beginnersmind

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

- triciaal

|dw:1441581658688:dw|

- 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.

- dan815

yes that is right

- dan815

well instead of positions and stuff

- 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

- beginnersmind

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

- beginnersmind

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

- beginnersmind

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

- triciaal

|dw:1441581910685:dw|

- beginnersmind

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

- dan815

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

- beginnersmind

Not sure yet.

- beginnersmind

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

- beginnersmind

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

- triciaal

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

- triciaal

or like @beginnersmind calls it "doublecounting"

- 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 :)

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