Question

Consider integer solutions to the equation:
\(x_1+x_2+x_3+x_4+x_5=10\)
How many non-negative integer solutions are there such that \(x_1>x_2\)?

Answer 1

a dumb method wud be to consder cases
x1 - x2 = 1,2,3,4,5,6,7,8,9

Answer 2

@Zarkon help

Answer 3

use the previous answer and symmetry

Answer 4

you need more hints?

Answer 5

No. of non-negative integer solutions without restrictions = C(14,4) = 1001
No. of non-negative integer solutions with either \(x_1>x_2\) or \(x_2>x_1\) = C(14,4) - 161 = 1001-161 = 840
If you say use symmetry, I suppose no. of non-negative integer solutions with \(x_1>x_2\) = no. of non-negative integer solutions with \(x_1<x_2\)?!
So, we get no. of non-negative integer solutions with \(x_1>x_2\) = 840/2 = 420?
Ehhhh....

Answer 6

correct

Answer 7

beautiful ! XD

Answer 8

Thanks, but how can you explain that symmetry concept?

Answer 9

you already did.


there are 3 possibilities \(x_1>x_2,x_1=x_2,x_1<x_2\)

by the symmetry of the problem \(x_1>x_2x_1<x_2\) have to have the same number of possibilities (you could switch the roles of \(x_1\) and \(x_2\) and you would still have the same number f solutions)

Answer 10

or you could do it the long way

\[\sum_{i=0}^{10}\sum_{j=i+1}^{10-i}{2+10-i-j\choose 2}=420\]

;)

Answer 11

typo

\[\sum_{i=0}^{4}\sum_{j=i+1}^{10-i}{2+10-i-j\choose 2}=420\]

Answer 12

that would be going through the cases

Answer 13

@RolyPoly , how would the total number of non-negative integer solutions without restriction be C(14,4)? I understand that C(14,4) means 14 choose 4, but I am curious where that came from.

Answer 14

Stars and bars :D

Answer 15

:)

Answer 16

https://brilliant.org/assessment/techniques-trainer/stars-and-bars/

Answer 17

Okay, give me a moment to chew on that. I am familiar with stars and bars.

Answer 18

@ganeshie8 , thanks for the link -- I am enjoying reading!

Answer 19

So, for the symmetry, after eliminating the solutions with \(x_1 = x_2\), we only have \(x_1>x_2\) or \(x_2>x_1\). Since we can switch the role of \(x_1\) and \(x_2\) without affecting the values of other variables, the number of solutions with \(x_1>x_2\) and with \(x_2>x_1\) are the same.
Does this explanation looks good? Or still missing something?

Answer 20

that looks fine

Answer 21

you can think of it like this too..


you have x1 and x2...but I called them x2 and x1 (backwards from you) so even though we are using a different order ...we are really doing the same problem.

Answer 22

Got it! Thanks!! :D