Question

In how many different ways three persons A, B, C having 6, 7 and 8 one rupee coins respectively can donate Rs.10 collectively?

Answer 1

@shubhamsrg

Answer 2

Can any one of them choose not to give anything?

Answer 3

can some give more and some give less??

Answer 4

@sambhav__jain can give max 6, b can give max 7, c can give max 8.
i think so @terenzreignz

Answer 5

Then, let's assume A gives 0.
The problem becomes how many ways B and C can give 10.

Answer 6

are you sure you can assume a gives nothing?

Answer 7

We can't. But we can consider all the cases where A gives 0, A gives 1, etc.

Answer 8

well that would be quite long. but how would you solve if b and c give10?

Answer 9

Well, there are six ways, depending on how many B gives.
B gives 7, C gives 3
B gives 6, C gives 4
B gives 5, C gives 5
B gives 4, C gives 6
B gives 3, C gives 7
B gives 2, C gives 8

B gives 1....C can't give enough... B can't give just 1.
Six possibilities if A gives 0

Now check if A gives 1. The problem is how many ways B and C can give 9

Answer 10

hmmm. got it. but this would be too long for an objective question.

Answer 11

firstly you may wanna read this:
http://www.cuemath.com/2012/permutations-and-combinations-counting-the-number-of-solutions-of-an-integral-equation/

let A give x, B gives y and C gives z
then
x+y+z = 10

where x<=6
y<=7
z<=8

--> 6-x >= 0
7-y >=0
8-z >=0

our expression is
x+y+z=10 , we have to find its integer solutions.

=> (6-x) + (7-y) + (8-z) = (6+7+8)-10

(6-x) + (7-y) + (8-z) = 11

hence our ans will be directly C(11+3-1 ,3-1) = C(13,2) = 78 ways

Hope that helped ?

Answer 12

yes i did that too. but the answer is 47.
i found this solution...
Suppose they are A B C
now A+B+C=10
=> So the number of integral solution =12C2
Now A cannot give more than 6 so the cases where A gives 7 or more need to be deducted
A=A'+7
=> So A'+B+C=3 the number of ways = 5C2.
=> Same way for B it's 4C2 and for C it's 3C2
=> Total = 12C2 - (5C2 + 4C2 + 3C2) = 47

why did they do A=A'+7?

Answer 13

Sorry had got dc.
Well yes there is a slight flaw in my method.
(6-x) + (7-y) + (8-z) = 11

The formula is incorrectly applied here as it does not acomodate for 6-x>=0, according to it, it treats it as X and this will include X=7, even though that'll make x negative

hmm
we need to subtract cases from 78
1st case : x>=6 --> 6C2
2nd case: y>=7 --> 5C2
and 3rd case : 4C2

final ans= 78 - 6C2 -5C2 - 4C2 = 47

Answer 14

am sorry that'll be for x>=7, y>=8 and z>=9

Answer 15

i am sorry.
"The formula is incorrectly applied here as it does not acomodate for 6-x>=0, according to it, it treats it as X and this will include X=7, even though that'll make x negative"
didnt get that. :O

Answer 16

@shubhamsrg

Answer 17

@Mashy

Answer 18

combinations ahhh! :P

Answer 19

not your forte? :P

Answer 20

not really.. but i know the concepts.. but m not getting a lead to start it :P

Answer 21

all that i dont understand here is this...
"The formula is incorrectly applied here as it does not acomodate for 6-x>=0, according to it, it treats it as X and this will include X=7, even though that'll make x negative"
i dont understand what 'shubhang' meant. :P

Answer 22

can we do it this way

case 1.. B and C give 4 rupee together.. so A has to give 6
work out how many ways this is possible
case 2 .. B and C give 5 rupee togther.. so A has to give 5

.. last case.. B and C give 9 rupee togther so A has to give 1 :D

Answer 23

you want me to count 47 ways like that? :P

Answer 24

no.. only 6 cases like that :-/

see case 1.. you have to find in how many ways can B and C give 4 rupee together :D

Answer 25

let me try and work out!

Answer 26

hmm. thats pretty much what terenz said.
i like shubhams method, but i cant understand it. :P

Answer 27

so case 1.. B and C can do it as 1,3 2,2 3,1.... hence case 1 is 3
case 2.. B anc C give 5 rupee together.. so 1,4... 2,3...3,2....4,1... hence case 2 is 4
case 3......................6...... togther so 1,5....2,4...3,3...4,2....5,1.. hence case 3 is 5

case 4 ... is similary 6
case 5 is 7
case 6 8

so the answer is 8+7+6+5+4+3 and thats 33.. where did i go wrong? :O

Answer 28

are we considering zero as well?? not giving as well?

Answer 29

yup.

Answer 30

ohh.. then
Case 1 .. 5
case 2 ..6
case 3.. 7
case 4..8
case 5..9
case 6..10
and one more case (A =0) .. 11

wait now m getting more how? :O .. i just added 2 to each case :-/

Answer 31

oh damn.. forget it.. i suck at this!

Answer 32

haha. i have to go now. i'll come back and have a look. :P

Answer 33

me going too bye :P

Answer 34

@yrelhan4 ,, terenzreignz has a point there

Answer 35

it leads to the solution

Answer 36

sure he has. yes it leads to the solution. but objective questions generally dont have long solutions.

Answer 37

Will anyone tell me from where did "shubhang" actually came into existence? -_-
"The formula is incorrectly applied here as it does not accommodate for 6-x>=0, according to it, it treats it as X and this will include X=7, even though that'll make x negative"

We have to deal with
(6-x) + (7-y) + (8-z) = 11

or
X + Y + Z =11
where X=6-x and Y=7-y and Z=8-z

the formula works for X+Y+Z = 11 , with eqch >=0, i.e. the formulla will also take in cases where X=7 and above
but X can't be 7,it can be max 6, if its 7, it'll make x -ve.

Sorry my eng is not very good :/

Answer 38

hmm. seems i got it. thank you.
and i have no idea where shubhang came from. ask @Mashy ! :P

Answer 39

hmm, glad to help.