Question

Obtain the values of the following:

(1) ¹¹C₄

(2) ⁹C₀

Answer 1

(1) ¹¹C₄

ⁿC_r = \(\frac{n!}{r!(n−r)!}\)

∴ ¹¹C₄ = \(\frac{11!}{41(11−4)!}\)

= \(\frac{11×10×9×8×7!}{4×3×2×1×7!}\)

= \(\frac{7920}{24}\) = 330

(2) ⁹C₀

ⁿC₀ = 1

∴ ⁹C₀ = 1