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Prove that C(50,3) + C(50,4) = C(51,4)

Sagot :

By definition of the binomial coefficient,

[tex]C(n,k) = \dfrac{n!}{k!(n-k)!}[/tex]

so we have

[tex]C(50,3) + C(50,4) = \dfrac{50!}{3!47!} + \dfrac{50!}{4!46!} \\\\ = \dfrac{50!}{3!46!} \left(\dfrac1{47} + \dfrac14\right) \\\\ = \dfrac{50!}{3!46!} \times \dfrac{51}{188} \\\\ = \dfrac{51!}{3!46!} \times \dfrac1{4\times47} \\\\ = \dfrac{51!}{4!47!} \\\\ = \dfrac{51!}{4!(51-4)!} = C(51,4)[/tex]

as required.

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