Question regarding permutations and combinations?

Question

Hi, I was just wondering on how you are supposed to approach this question. I keep getting 114 as an answer, but the answers say it is 174. How would anyone do this question, because I feel like I'm not ever doing the permutation and combination questions correctly.

The way I did it was, I assigned symbols for gates for exact money (E) and gates for change (C). With the three cars stated in the question, one will require change, and the other two have exact change. So I assumed the order could be EEC, ECE or CEE.

EEC = 5*4*3 possible gates ECE = 5*2*3 CEE = 2*4*3

and adding them all up, I get 114.

Thanks

Answer 1

Ford can go in 2, Holden and Toyota can go in all 5 So two times [something] Since the Holden and Toyota can go in either of the other four (different gates). So you have... Holden driver can go in 4 gates, Toyota can go in 3 gates..

24 options if you don't care about order

But I think you need to care about which order they go in, so then you need to times it by 3 factorial = 6

So the answer I got is 2*4*3*3*2 = 144

That isn't 174 though?

Answer 2

I think the logic for the answer of 174 is that of.

All possible ways of cars driving through gates minus the possibility that cannot happen due to the conditions stated.

So with 5 gates and three cars.

That's (5x4x3) multiplied by order of entry 3 is 180. All possible permutations with order.

The instances where the conditions will not be met is the permutation where the two cars that have exact fare hog the change giving booths.

That's car a and car b in the change giving booths with three options for car c to take the wrong booth. Multiplied by 2 to switch car a and b(new permutation) giving 6.

So 180-6 =174

Answer 3

Another way of solving it is going through the following tedious analysis.

Let the Ford be A and the other two cars be b and c. We use different symbols to show how car b goes through these pay stations:-

1. {b}--- It uses the exact change gate on purpose.

2. (b) ---It does not use the exact change gate.

3. b ---A don’t care condition.

Judging from the arrival times, we have 6 different situations, namely Abc; Acb; bAc; cAb; bcA; and cbA. We will study each of these cases one by one.

For the Abc case, we can have the following 2 sub-cases:-

1.1 A{b}(c) -------- no. of ways = 2*1*3 = 6

1.2 A(b)c ------- = 2*3*3 = 18

The Acb case is exactly the same as above. And therefore, there are 24 ways.

For the bAc case. we can have the following 2 sub-cases:-

3.1 {b}A(c) -------- = 2*1*3 = 6

3.2 (b)Ac -------- = 3*2*3 = 18

The cAb case is exactly the same as above. And therefore, there are 24 ways.

For the bcA case, we can have the following 4 sub-cases:-

5.1 {b}{c}A ------------ = 0

5.2 {b}(c)A ----------- = 2*3*1 = 6

5.3 (b){c}A ---------- = 3*2*1 = 6

5.4 (b)(c)A ---------- = 3*2*2 = 12

The cbA case is exactly the same as above. And therefore, there are 24 ways.

The answer is then 6*24 = 144.