91Ó°ÊÓ

We have ten objects, and we want to make groups considering at least three objects at a time. We will use the combination formula to solve the problem: \(C(n,r)=\frac{n!}{r!(n-r)!}\) where n is the number of objects (10 in this case) and r is the number of objects to be chosen (minimum 3 in this case), and ! stands for the factorial operation.

Let's start calculating combinations for each group size: For choosing 3 objects: \(C(10,3) = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = 120\) For choosing 4 objects: \(C(10,4) = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = 210\) For choosing 5 objects: \(C(10,5) = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!} = 252\) For choosing 6 objects: \(C(10,6) = \frac{10!}{6!(10-6)!} = \frac{10!}{6!4!} = 210\) For choosing 7 objects: \(C(10,7) = \frac{10!}{7!(10-7)!} = \frac{10!}{7!3!} = 120\) For choosing 8 objects: \(C(10,8) = \frac{10!}{8!(10-8)!} = \frac{10!}{8!2!}=45\) For choosing 9 objects: \(C(10,9) = \frac{10!}{9!(10-9)!} = \frac{10!}{9!1!} = 10\) For choosing 10 objects: \(C(10,10) = \frac{10!}{10!(10-10)!} = \frac{10!}{10!0!} = 1\)

Now, we just need to add up the number of combinations for each group size to get the total number of groups: Total number of groups = 120 + 210 + 252 + 210 + 120 + 45 + 10 + 1 = 968 The total number of groups that can be formed from 10 objects taking at least three at a time is 968.