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I found them in a test but just able to solve and there are no answers announced, so please check my answer for

$$I(2,n)\ ,\ I(3,n)\ ,\ I(5,3)$$ given $$I(m,1) =I(1,n) =1\ and\ I (m+1,n) +I (m,n+1) =I (m+1,n+1)$$

These are my answer $$I(2,n) = {n}$$ $$I(3,n) = \frac{n(n+1)}{2}+1$$ $$I(5,3) = 15$$

youngboy
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1 Answers1

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$$I(m,n) = \binom {n+m-2}{n-1} = \frac {(n+m-2)!}{(n-1)!(m-1)!},$$ so the first and the third answers are correct while the second is not. $I(3,n)=\frac {n(n+1)} 2$.

John McClane
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  • $$I(3,n) = I(3,n-1) + I(2,n)\ (and\ I(2,n)=n)$$ Then it continues this pattern $$I(3,n) = I(3,n-2) + n-1$$ $$I(3,n) = I(3,n-3) + n-2$$ and I got $$I(3,n) = I(3,n-(n-1)) + n + (n-1) + (n-2) + ...+ n-(n-1)$$ $$I(3,n) = I(3,1) + \frac{n(n+1)}{2}$$

    what is my mistake?

     – youngboy Jun 19 '18 at 05:31
  • There are some mistype in 3rd and 4th lines For 3rd, $$I(3,n-1)\ not\ I(3,n)$$ and 4th is $$I(3,n-2)\ not\ I(3,n)$$ – youngboy Jun 19 '18 at 05:56
  • Mistake is here: $I(3,n) = I(3,n-(n-1)) + n + (n-1) + (n-2) + ...+ n-(n-1)$. It must be $I(3,n) = I(3,n-(n-1)) + n + (n-1) + (n-2) + ...+ n-(n-2)$. – John McClane Jun 20 '18 at 08:20