Question

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 ?

Answer 1

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 ?

Answer 2

so 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

Answer 3

if A gets 3
B can have between 1 pen and (12-3)=9 pens
if A gets 4
B can have between 1 pen and (11-3)=8 pens
if A gets 5
B can have between 1 pen and (10-3)=7 pens
if A gets 6
B can have between 1 pen and (9-3)=6 pens

Answer 4

u mean B=remaining 4 students

Answer 5

as an students A,B,C,D,E

Answer 6

hey but this listing all ABCDE method is time taking

Answer 7

so is this right 9!+8!+7!+6! ?
or aka 408960
?

Answer 8

425 is given

Answer 9

err I'm bad at counting sometimes
let me rethink

Answer 10

are u considering all pens are distinct or equal

Answer 11

\[{12\choose4}-{8\choose 4}\]

Answer 12

@zarkon this might be a dumb question
can you tell me how you got that

Answer 13

like your thinking and stuff
so i can have your counting wisdom

Answer 14

or do you understand @mathmath333 ?

Answer 15

stars and bars twice

Answer 16

i did'nt understand his logic

Answer 17

first he ignored the condition that A has 3 to 6 pens then he subtracted something

Answer 18

*-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

Answer 19

I ignored "at most 6 pens" then I subtracted the ones where there were 7 or more

Answer 20

r u still typong

Answer 21

b+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

Answer 22

The thing we used was this theorem:
\[\text{ For any pair of positive integers }n \text{ and } k, \text{ the number of distinct } \\ k-\text{tuples of positive integers whose sum is } n \text{ is given by } \left(\begin{matrix}n-1 \\ k-1\end{matrix}\right)\]

Answer 23

I used this theorem for each of those 4 equations I wrote

Answer 24

Thanks @zarkon for bringing up the stars and bars method.

Answer 25

thnks

Answer 26

give 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)

\[{12\choose 4}-{8\choose4}=425\]