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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 ?
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 ?

This Question is Closed

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & \normalsize \text{A teacher has to distribute 15 pens among 5 of his students such}\hspace{.33em}\\~\\ & \normalsize \text{that student A gets at least 3 and at most 6 pens.In how many }\hspace{.33em}\\~\\ & \normalsize \text{ ways can this be done ?}\hspace{.33em}\\~\\ & a.)\ 495 \hspace{.33em}\\~\\ & b.)\ 77 \hspace{.33em}\\~\\ & c.)\ 417 \hspace{.33em}\\~\\ & d.)\ 435 \hspace{.33em}\\~\\ \end{align}}\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0only student A gets atleast 3 to 6 pens necessarily

dan815
 one year ago
Best ResponseYou've already chosen the best response.2lets do 4 cases then Student A 3 4 5 6

dan815
 one year ago
Best ResponseYou've already chosen the best response.2when student A gets 3 pens we have 12 pens to distribute among 4 students

dan815
 one year ago
Best ResponseYou've already chosen the best response.2why are you doing 12 choose 4

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0cuz i left with 4 students and 12 pens

dan815
 one year ago
Best ResponseYou've already chosen the best response.2wait umm from your previous questions, when they ask questions like this does that mean the pens are distinguishable or not?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i think pens are similar/identical

dan815
 one year ago
Best ResponseYou've already chosen the best response.2okay phew that makes it a lot simpler

dan815
 one year ago
Best ResponseYou've already chosen the best response.2so we have 12 pens and 4 students think about it like stars and bars question

dan815
 one year ago
Best ResponseYou've already chosen the best response.2now can u do the other cases?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.015C3+14C3+12C3+11C3 ?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0u mean it is correct

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0it is 1325 not in options

dan815
 one year ago
Best ResponseYou've already chosen the best response.2hm thats weird, i dont see anything wrong with what we did, unless the question should be interpretated differently

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0what if the pens are distinct .

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2You had 15 pens originally but you gave at least 3 to A.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2then it iwll be even more

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2So the highest case is 12C3

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2and the lowest 9C3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.012C4+11C4+10C4+9C4 is wrong ?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0answer given is d.) 435

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2@Halmos Yes, because you're not choosing people. You are choosing where to put the dividing line between people's stack of pens.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441580705091:dw orginal credit to dan. You started out with 15 pens. But you gave 3 to A. So you have 12 left.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2right but after u get 12 pens then u got a total of 15 spots from the 12 pens and 3 divider lines

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2BTW 12C3 is incorrect too. It should be 11C3. Because there's 11 positions to put your dividing line.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441580893991:dw Or 13. And you can put 2 dividing lines in the same position as well :(

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2So it's really 3^13 +3^12 +3^11 +3^10

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i reallly think its 15 choose 3 for dividing 12 pens between 4ppl if we allow someone to get 0 pens

dan815
 one year ago
Best ResponseYou've already chosen the best response.2for 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
 one year ago
Best ResponseYou've already chosen the best response.2which is true 0 ,5 1, 4 2, 3 3,2 4,1 5, 0

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1student A at least 3 pens 3 or 4 or 5 or 6 next student 12 for 4 students dw:1441581202316:dw

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2So how do you get 15 places from 12 objects?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2what we are about is the order of these 15 things, that determines how its been split up

dan815
 one year ago
Best ResponseYou've already chosen the best response.2like 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
 one year ago
Best ResponseYou've already chosen the best response.2so in this case u have 5 spots for your 2 bars, 5C2

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2You are confounding choosing positions for your dividers and arranging objects in a certain order.

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441581658688:dw

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2The 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
 one year ago
Best ResponseYou've already chosen the best response.2well instead of positions and stuff

dan815
 one year ago
Best ResponseYou've already chosen the best response.2you 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
 one year ago
Best ResponseYou've already chosen the best response.2That's another way to do it, yes. They should give the same answer. Mine is 3^13, yours is 15!/(12!*3!)

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2Correction Mine is (3^13)/3!, yours is 15!/(12!*3!)

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.215!/(12!*3!) = 15C3 Mind Blown

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441581910685:dw

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2Ok, @dan815, finally figured it out. Your method works, mine doesn't. Sorry, I'm slow tonight.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ah its all good, so why do u think its not in the answers though

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2I'm trying to figure it if there's any double counting going on but I can't see any.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2Alright, I give up. Good luck everyone. @mathmath33 you might want to ask your teacher.

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1isn't Student A one particular student of the 5? if a student then you have multiples

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1or like @beginnersmind calls it "doublecounting"

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2Well, 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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