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 ?

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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 ?

Mathematics
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\(\large \color{black}{\begin{align} & \normalsize \text{A teacher has to distribute 15 pens among 5 of his students such}\hspace{.33em}\\~\\ & \normalsize \text{that student A gets at least 3 and at most 6 pens.In how many }\hspace{.33em}\\~\\ & \normalsize \text{ ways can this be done ?}\hspace{.33em}\\~\\ & a.)\ 495 \hspace{.33em}\\~\\ & b.)\ 77 \hspace{.33em}\\~\\ & c.)\ 417 \hspace{.33em}\\~\\ & d.)\ 435 \hspace{.33em}\\~\\ \end{align}}\)
only student A gets atleast 3 to 6 pens necessarily
lets do 4 cases then Student A 3 4 5 6

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when student A gets 3 pens we have 12 pens to distribute among 4 students
12C4=495
u cant just do that
why are you doing 12 choose 4
cuz i left with 4 students and 12 pens
wait umm from your previous questions, when they ask questions like this does that mean the pens are distinguishable or not?
i think pens are similar/identical
okay phew that makes it a lot simpler
so we have 12 pens and 4 students think about it like stars and bars question
15C3
|dw:1441579954826:dw|
yes 15 choose 3
now can u do the other cases?
14C3 ?
go on
15C3+14C3+12C3+11C3 ?
yeah
i m dumb
u mean it is correct
it is 1325 not in options
hm thats weird, i dont see anything wrong with what we did, unless the question should be interpretated differently
what if the pens are distinct .
You had 15 pens originally but you gave at least 3 to A.
then it iwll be even more
So the highest case is 12C3
and the lowest 9C3
12C4+11C4+10C4+9C4 is wrong ?
why 12C3?
answer given is d.) 435
yea why 12C3
@Halmos Yes, because you're not choosing people. You are choosing where to put the dividing line between people's stack of pens.
|dw:1441580705091:dw| orginal credit to dan. You started out with 15 pens. But you gave 3 to A. So you have 12 left.
right but after u get 12 pens then u got a total of 15 spots from the 12 pens and 3 divider lines
BTW 12C3 is incorrect too. It should be 11C3. Because there's 11 positions to put your dividing line.
|dw:1441580893991:dw| Or 13. And you can put 2 dividing lines in the same position as well :(
So it's really 3^13 +3^12 +3^11 +3^10
i reallly think its 15 choose 3 for dividing 12 pens between 4ppl if we allow someone to get 0 pens
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
which is true 0 ,5 1, 4 2, 3 3,2 4,1 5, 0
student A at least 3 pens 3 or 4 or 5 or 6 next student 12 for 4 students |dw:1441581202316:dw|
So how do you get 15 places from 12 objects?
3 for divisions
what we are about is the order of these 15 things, that determines how its been split up
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
so in this case u have 5 spots for your 2 bars, 5C2
You are confounding choosing positions for your dividers and arranging objects in a certain order.
|dw:1441581658688:dw|
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.
yes that is right
well instead of positions and stuff
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
That's another way to do it, yes. They should give the same answer. Mine is 3^13, yours is 15!/(12!*3!)
Correction Mine is (3^13)/3!, yours is 15!/(12!*3!)
15!/(12!*3!) = 15C3 Mind Blown
|dw:1441581910685:dw|
Ok, @dan815, finally figured it out. Your method works, mine doesn't. Sorry, I'm slow tonight.
ah its all good, so why do u think its not in the answers though
Not sure yet.
I'm trying to figure it if there's any double counting going on but I can't see any.
Alright, I give up. Good luck everyone. @mathmath33 you might want to ask your teacher.
isn't Student A one particular student of the 5? if a student then you have multiples
or like @beginnersmind calls it "doublecounting"
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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