Applied Algebra (Prove the identity)?

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SUMMARY

The forum discussion centers on proving the combinatorial identity: sum (i from 0 to k) { C(m, i) * C(n, k - i) } = C(m + n, k) for positive integers m and n. The hint provided suggests utilizing the polynomial equation sum (k from 0 to m+n) {C(m + n, k) * z^k } = (1 + z)^(m+n) = ((1+z)^m) * ((1+z)^n). Participants are encouraged to express (1+z)^m and (1+z)^n as sums over powers of z using binomial coefficients and equate the powers of z on both sides to establish the identity.

PREREQUISITES
  • Understanding of binomial coefficients, specifically C(m, i).
  • Familiarity with polynomial equations and their expansions.
  • Knowledge of combinatorial identities and their proofs.
  • Basic algebraic manipulation skills.
NEXT STEPS
  • Study the properties of binomial coefficients and their applications in combinatorics.
  • Learn about polynomial expansions and the Binomial Theorem.
  • Explore combinatorial proofs and techniques for proving identities.
  • Investigate generating functions and their role in combinatorial mathematics.
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Mathematicians, students studying combinatorics, and anyone interested in algebraic proofs and identities will benefit from this discussion.

chredhat
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let m,n be positive integer. Prove the identity:

sum (i from 0 to k): { C(m, i) * C(n, k - i) } = C(m + n, k)

Hint: Consider the polynomial equation:

sum (k from 0 to m+n) {C(m + n, k) *z^k } = (1 + z)^(m+n) = ((1+z)^m) * ((1+z)^n)

I tried long time, still have no idea.
 
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Express (1+z)^m and (1+z)^n as sums over powers of z using the binomial coefficients as they did in the hint for (1+z)^(m+n). Now equate equal powers of z on both sides.
 

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