In how many ways can 10 people be divided into two groups, one with 7 people and the other with 3 people?

In terms of calculating, it's easier to pick 3 people to put in the smaller group and then just put the rest in the group of 7.

C(10,3) * C(7,7)
= C(10,3) * 1
= (10 * 9 * 8) / (3 * 2 * 1)
= 720/6
= 120 ways

Another way to solve this is:
10! / (7! 3!)
= 120 ways

In general, if you have N people and you are dividing them into groups of size A, B, C, etc. then the formula is:N! / (A! B! C!...)

The one gotcha is if some of the groups have the same size, then you need to consider the number of arrangements of those groups. For example, if you had 15 people and you were dividing them into groups of 5, the groups are essentially indistinguishable. So you need to further divide by 3! for the ways that you could shuffle the 3 groups (of 5) and still have the same set of people chosen for each specific group. So in this example it would be: 15! / (5! 5! 5! * 3!)

And if you had more duplicates, you'd divide further. So if you had 23 people divided into 3 groups of 5 and 2 groups of 4, then it would be: 23! / (5! 5! 5! 4! 4! * 3! 2!)

But for this simple case, it is just C(10,3) = 120 by choosing the people in the group of 3 and putting everyone else in the group of 7.

Answer:
120 ways

From the 10 people, select 7 people into the first group.
Put the rest 3 people into the second group.

Number of ways
= ₁₀C₇ × ₃C₃
= 120 × 1
= 120