Question regarding permutations and combinations? 
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

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:-


*

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

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

*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.
