# How many seating arrangements are possible?

There are 7 employees (E) and 5 non-employees (NE). Four of them must be chosen to ride in a four-seat vehicle. How many seating arrangments are possible if at least two passengers must be non-employes and only an employee can be the driver.

This is what I did:

The car has 4 seats: Right top (RT), left top (LT), right bottom (RB), left bottom (LB) - from a high-view perspective.

Case 1: RT: Employee, LT: Non-Employee, RB: Employee, LB: Non-Employee.

Hence, 7*5*6*4 = 840

Case 2: RT: Employee, LT: Non-Employee, RB: Non-Employee, LB: Employee.

Hence, 7*5*4*6 = 840

Case 3: RT: Employee, LT: Employee, RB: Non-Employee, LB: Non-Employee.

Hence, 7*6*5*4 = 840

Case 4: RT: Employee, LT: Non-Employee, RB: Non-Employee, LB: Non-Employee.

Hence, 7*5*4*3 = 420

840 + 840 + 840 + 420 = 2940.

So there are 2940 seating arrangments. Is my logic wrong?

• ways of choosing $$1$$ employee an $$3$$ non-employees: $$\color{blue}{7\binom{5}{3}}$$
• ways of seating them: $$\color{blue}{3!}$$
• ways of choosing $$2$$ employee an $$2$$ non-employees: $$\color{green}{\binom{7}{2}\binom{5}{2}}$$
• ways of seating them:$$\color{green}{\underbrace{2}_{driver}\cdot 3!}$$
$$\color{blue}{7\binom{5}{3}}\cdot \color{blue}{3!} + \color{green}{\binom{7}{2}\binom{5}{2}}\cdot \color{green}{2\cdot 3!} = 2940$$