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

@hartnn :D

Answer 2

say, \(x_1 = x_2 = x\)

\(2x + x_3 + x_4 + x_5 = 10\)

\(x_3 + x_4 + x_5 = 2(5-x)\)

\(x = 0,1,2,3,4\)

Answer 3

add all 5 cases

Answer 4

Hmm... Why can't x be 5?

Answer 5

If x=5, we have \(x_3=x_4=x_5 = 0\)

Answer 6

oh yah it can be 5 also thats trivial case
x3 = x4=x5 = 0

Answer 7

Hmmm
\(x_3+x_4+x_5=2(5−x)\)
When x=0:
\(x_3+x_4+x_5=10\)
No of solutions = C(12, 2) = 66
When x=1:
\(x_3+x_4+x_5=8\)
No of solutions = C(10, 2) = 45
When x=2:
\(x_3+x_4+x_5=6\)
No of solutions = C(8, 2) = 28
When x=3:
\(x_3+x_4+x_5=4\)
No of solutions = C(6, 2) = 15
When x=4:
\(x_3+x_4+x_5=2\)
No of solutions = C(4, 2) = 6
When x=5:
\(x_3+x_4+x_5=0\)
No of solutions = C(2, 2) = 1

Total number of solutions = 66+45+28+15+6+1=161?!

Answer 8

\(\binom{12}{2} + \binom{10}{2} +\binom{8}{2} +\binom{6}{2} +\binom{4}{2} + 1\)

Answer 9

http://www.wolframalpha.com/input/?i=%5Cbinom%7B12%7D%7B2%7D+%2B+%5Cbinom%7B10%7D%7B2%7D+%2B%5Cbinom%7B8%7D%7B2%7D+%2B%5Cbinom%7B6%7D%7B2%7D+%2B%5Cbinom%7B4%7D%7B2%7D

add 1

Answer 10

Thanks! Coming up next: Number of solutions again!! :D

Answer 11

hah sure, these are fun if we know stars and bars !!

Answer 12

We know stars and bars :D