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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 and each remaining
student gets at least one pen.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 and each remaining student gets at least one pen.In how many ways can this be done ?

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1A teacher has to distribute \(15\) pens among \(5\) of his students such that student A gets at least \(3\) and at most \(6\) pens and each remaining student gets at least one pen.In how many ways can this be done ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so A can get 3,4,5, or 6 pens if A gets 3 pens then there are 12 pens left to distribute if A gets 4 pens then there are 11 pens left to distribute if A gets 5 pens then there are 10 pens left to distribute if A gets 6 pens then there are 9 pens left to distribute

freckles
 one year ago
Best ResponseYou've already chosen the best response.2if A gets 3 B can have between 1 pen and (123)=9 pens if A gets 4 B can have between 1 pen and (113)=8 pens if A gets 5 B can have between 1 pen and (103)=7 pens if A gets 6 B can have between 1 pen and (93)=6 pens

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1u mean B=remaining 4 students

freckles
 one year ago
Best ResponseYou've already chosen the best response.2as an students A,B,C,D,E

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1hey but this listing all ABCDE method is time taking

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so is this right 9!+8!+7!+6! ? or aka 408960 ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2err I'm bad at counting sometimes let me rethink

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1are u considering all pens are distinct or equal

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.3\[{12\choose4}{8\choose 4}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2@zarkon this might be a dumb question can you tell me how you got that

freckles
 one year ago
Best ResponseYou've already chosen the best response.2like your thinking and stuff so i can have your counting wisdom

freckles
 one year ago
Best ResponseYou've already chosen the best response.2or do you understand @mathmath333 ?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i did'nt understand his logic

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1first he ignored the condition that A has 3 to 6 pens then he subtracted something

freckles
 one year ago
Best ResponseYou've already chosen the best response.2*hmmm... so we have b+c+d+e=9 and b+c+d+e=8 and b+c+d+e=7 and b+c+d+e=6 so stars and bars 4 times? \[4(8 C 3 +7 C3 + 6 C 3+5 C3)\] hmmm but we want to subtract out any solutions having 0 in it ... lol

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.3I ignored "at most 6 pens" then I subtracted the ones where there were 7 or more

freckles
 one year ago
Best ResponseYou've already chosen the best response.2b+c+d=4 2 bars 4 stars no bars at end or touching... these count as zeros... **** **** **** so I think there are 3 ways here... b+c+d=5 ***** ***** ***** ***** ***** ***** 6 ways? b+c+d=6 ****** ****** ****** ****** ****** ****** ****** ****** ****** ****** 10 ways?  3 ways=(3 choose 2) 6 ways=(4 choose 2) 10 ways=(5 choose 2) ... hmm we want oh... thanks to @ganeshie8 I think I have this using also the stars and bars.. By the way the earlier things I wrote on this post was for me. Thought maybe you would want to see the thinking or whatever... so we have the equations: b+c+d+e=12 b+c+d+e=11 b+c+d+e=10 b+c+d+e=9 so finding the number of positive solutions to each of those gives us: \[11C3+10C3+9C3+8C3=425\] this does give us the right answer

freckles
 one year ago
Best ResponseYou've already chosen the best response.2The thing we used was this theorem: \[\text{ For any pair of positive integers }n \text{ and } k, \text{ the number of distinct } \\ k\text{tuples of positive integers whose sum is } n \text{ is given by } \left(\begin{matrix}n1 \\ k1\end{matrix}\right)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I used this theorem for each of those 4 equations I wrote

freckles
 one year ago
Best ResponseYou've already chosen the best response.2Thanks @zarkon for bringing up the stars and bars method.

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.3give 3 pens to student A and 1 pen to the other students that leaves 8 pens. use stars and bars on 8 pens and 5 students \[{8+4\choose 4}={12\choose 4}\] now give 7 pens to A then we have \[{4+4\choose 4}={8\choose 4}\] (we don't want these in our total count so we subtract) \[{12\choose 4}{8\choose4}=425\]
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