How to Prove This Combinatorial Identity Involving Binomial Coefficients?

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The discussion focuses on proving the combinatorial identity C(n+r+1, r) = C(n+r, r) + C(n+r-1, r-1) + ... + C(n, 0) and its alternative form involving C(n+k, n). Participants emphasize the need to utilize the definition of binomial coefficients, C(n, r) = n!/[r!(n-r)!], to identify patterns and potential cancellations. Suggestions include exploring known combinatorial identities that may simplify the proof process. The conversation highlights the importance of demonstrating work to facilitate assistance in solving the problem. Ultimately, the goal is to establish the validity of the identity through combinatorial reasoning.
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Homework Statement


For positive integers n, r show that C(n+r+1, r) = C(n+r, r) + C(n+r-1, r-1) + ... + C(n+2, 2) + C(n+1, 1) + C(n, 0) = C(n+r, n) + C(n+r-1, n) + ... + C(n+2, n) + C(n+1, n) + C(n, n)


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The Attempt at a Solution

 
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You need to show some work before we can help with your homework. As a start, you could try stating the definitions of e.g. C(n+r+1, r).
 
Sorry. I know the definition of C(n, r) = n!/[r!(n-r)!] but I still can't figure out the pattern. Are things supposed to cancel out?...other identities that I'm forgetting and could help in this question?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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