210. she has 210 ways to choose 6 friends from the list of 10 to sit at the table closest to the head table no matter the order.
This is a problem of combinations and can be solved using the equation [tex]nC_{k}=\frac{n!}{k!(n-k)!}[/tex], where n! and k! is the factorial of a number. The factorial is defined in principle as the product of all positive integers from 1 (ie, natural numbers) to n.
She has a list of 10 friends and we want to know in how many ways she can choose 6 friends.
Using the combinations equation, with n = 10 and k = 6:
[tex]10C_{6}=\frac{10!}{6!(10-6)!}=\frac{10!}{6!(4!)}=\frac{10.9.8.7}{4.3.2.1}=\frac{5040}{24} =210[/tex]
