Question

find the value of 10C3 + 11C8 - 5C4

Answer 1

ⁿCₓ = $\frac{n!}{(n-r)!.r!}$

¹⁰C₃ = $\frac{10!}{(10-3)!.3!}$
= $\frac{10(9)(8)(7!)}{7!.3!}$
= $\frac{720}{6} = 120$

¹¹C₈ = $\frac{11!}{(11-8)!.8!}$
= $\frac{11(10)(9)(8!)}{3!.8!}$
= $\frac{11(10)(9)}{3(2)(1)}$
= $11(5)(3) = 165$

⁵C₄ = $\frac{5!}{(5-4)!.4!}$
= $\frac{5(4!)}{4!} = 5$

¹⁰C₃ + ¹¹C₈ - ⁵C₄ = 120 + 165 - 5
= 120 + 160
= 280

hope it helps
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