Eg 2. There are 10 seats in a row and three are empty. How many ways can this be arranged?
\[
\frac{10 !}{2 !}=604800
\]
Answer
\(\boxed{120}\) ways to arrange the 10 seats with 3 empty seats.
Step 1 ：Let's find the number of ways to arrange 7 occupied seats and 3 empty seats in a row. We can use the formula for combinations: \(C(n, k) = \frac{n!}{k!(n-k)!}\), where n is the total number of seats (10) and k is the number of empty seats (3).
Step 2 ：n = 10
Step 3 ：k = 3
Step 4 ：Calculate the combinations: \(C(10, 3) = \frac{10!}{3!(10-3)!}\)
Step 5 ：\(C(10, 3) = \frac{10!}{3!7!}\)
Step 6 ：\(C(10, 3) = \frac{3628800}{(6)(5040)}\)
Step 7 ：\(C(10, 3) = \frac{3628800}{30240}\)
Step 8 ：\(C(10, 3) = 120\)
Step 9 ：\(\boxed{120}\) ways to arrange the 10 seats with 3 empty seats.
