Discussion Problem_(23)

[duck]

Q1) The number of different solutions (x,y,z) of the equation x+y+z =10 is possible where each of x,y and z is a positive integer.

Q2) A salesman sold two pipes at Rs12 each. His profit on one was 20% and the loss on the other was 20%.
Then, on the whole, how much he gained/lost?

[Sinistral]

1) 9C2

2) 4% loss

[MI]

1. 9c2

2. loss of 4%

[Mauli]

i made cases and solved. though i got 36.. but how did u guys get 9C2. definitely u have done it with a shorter method.
please share it here.

[Sumit]

1).9C2
2).4% loss.

@Mauli: This is a simple question of distribution.
n Formula for distributing 'n' identical things in 'r' different boxes(or say person)=n+r-1Cr-1.
but notice this formula also provide those solutions in which a person can have zero thing(i.e nothing).
But the question given by duck say positive integer. So, in this case If we distribute 1 to each variable before solving...then we will left with x+y+z=7. Now you can apply above formula...

[duck]

Correct Answers:
1) 36
2) Loss of Re.1