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${}^{20}C_{4} + 2 {}^{20}C_{3} + {}^{20}C_{2} {}^{- 22}C_{18} is equal to$

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Correct option is A

C4 20+220C3+C2 20-C18 22=C4 21+C3 20+C2 20-C18 22 Cr n+Cr-1 n= n+1Cr =C4 21+C3 21-C18 22=C4 22-C18 22 =C18 22-C18 22=0.

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C4 20+2 C3 20+C2 20C18 -22 is equal to

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${}^{20}C_{4} + 2 {}^{20}C_{3} + {}^{20}C_{2} {}^{- 22}C_{18} is equal to$