388 views
1 votes
1 votes
number of non negative integer solution for

1)x1+x2+x3+x4+x5<=10<br />
2)x1+x2+x3+x4+x5<10

1 Answer

Best answer
0 votes
0 votes
1. Any solution of the equation can be seen as partitioning 10 (or less) indistinguishible objects into 5 groups (corresponding to x1,x2,x3,x4,x5), where each group may be empty (because solution is non-negative integer).

This can be solved by stars and bars (Theorem 2 on https://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29), in which number of stars is 10 (or less) here, and number of bars is 5.

For a particular number of $n$ stars and $k$ bars, solution is given by $\binom{n+k-1}{k-1}$.

Here $n$ varies from 0 to 10, and $k$ is 5 So total number of ways is $\sum_{n=0}^{10} \binom{n+5-1}{5-1}$.

2. This is same as 1st, except that $n$ varies from 0 to 9.

Related questions

0 votes
0 votes
1 answer
2
. asked Feb 23, 2017
623 views
The number of ways in which three non-negative integers can be chosen such that is
3 votes
3 votes
4 answers
4
pankaj9310 asked Jan 29, 2015
8,260 views
How many solutions are there to the equation x+y+z=17 ?They are non-negative integers A) 120 B)171 C)180 D)121