2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

You solution looks good but you may shorten it a bit using some combinatorics:

  • 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:$\color{green}{\underbrace{2}_{driver}\cdot 3!}$

All together:

$$\color{blue}{7\binom{5}{3}}\cdot \color{blue}{3!} + \color{green}{\binom{7}{2}\binom{5}{2}}\cdot \color{green}{2\cdot 3!} = 2940$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.