The correct answer is: 330
Explanation:
The formula for combination is given as follows:
\(^nC_r = \frac{n!}{r! (n-r)!}\) --- (1)
Where n = 11
r = 4
Plug in the values in equation (1) and calculate:
\(^nC_r = \frac{n!}{r! (n-r)!} \\
^{11}C_{4} = \frac{11!}{4! (11-4)!} \\
^{11}C_{4} = \frac{11*10*9*8*(7!)}{4! (7)!} \\
^{11}C_{4} = \frac{11*10*9*8}{4!} \\
^{11}C_{4} = \frac{11*10*9*8}{4*3*2*1} \\
^{11}C_{4} = 330 \\\)
Hence the correct answer is 330.
