Answer:
56
Step-by-step explanation:
Here, n = 8, r = 3
[tex] \huge \orange{ \because \: ^nC_r = \frac{n!}{r!(n - r)!} } \\ \\ \huge \therefore \: ^8C_3 = \frac{8!}{3!(8 - 3)!} \\ \\ \huge \therefore \: ^8C_3 = \frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 1 \times 5!} \\ \\ \huge \red{ \boxed{\therefore \:^8C_3 = \purple{56}}}[/tex]
