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only student A gets atleast 3 to 6 pens necessarily

lets do 4 cases then
Student A
3
4
5
6

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

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

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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

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

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

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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

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
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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

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.

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yes that is right

well instead of positions and stuff

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

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

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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"