anonymous
  • anonymous
How do I solve for m,n when f(m,n)=2?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
The given function was \[f\left ( m,n \right )=\binom{m+n-1}{2}+m\] and f(1,1)=1 How do I find what is m and n when f(m,n)=2?
anonymous
  • anonymous
the brackets meant "n choose r"
anonymous
  • anonymous
When you look at this thing, you end up with one equation in two unknowns. To solve it, I made some assumptions on the allowable values m and n could take (basically, m and n could be non-negative integers). Attached is what I worked out for the function after expanding the binomial coefficient. I organised it so that the result would be quadratic in m, for which I then solved. Then, since I assumed only non-negative integer solutions could be found, it meant only those n for which 8n+9 would be a perfect square, could be taken. From the assumption, then, the lowest n to test would be 0, leading to m = 2 or -1. The next n that would give a perfect square is 5, but for that choice, you end up with m = -1 or -8, and any greater n would lead to more negative solutions for m (since the negative growth in the linear factor outdoes any positive growth in the square root). So we can conclude n=0 and m=2. When you check this, you get\[f(2,0)=(m+n-1;2)+m = (2+0-1;2)+2=(1;2)+2=0+2=2\]as required. I hope this is what you want. Please let me know how it goes :)
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