These seemingly difficult permutation problems can be solved using some simple ideas.

The basic idea underlying solving any permutation is that we are multiplying together the number of possibilities for each slot. In the simplest case of filling 7 parking spaces with 7 different cars, there would be 7 possibilities for the first slot, 6 for the second, 5 for the third, and so on, but this is different. Still, solving it involves using the same basic idea.

We have two constraints. One is that that P and Q must be next to each other. The other is that S has to be to the right of both of them.

To handle the P and Q constraint I am going to make P and Q next to each other one unit. So now we are down to 6 units. However, that one unit has two versions, P to the left of Q and P to the right.

If we didn't have the S constraint we could do two permutations of 6 units. We would have 6 x 5 x 4 x 3 x 2 x 1 for P to the left of Q and the same thing for P to the right of Q. So 2(6!) would be our answer.

However we have the additional constraint of S having to be to the right of the P and Q unit.

One way to figure out how many arrangements have S to the right of P and Q would be to do multiple permutations starting with P and Q and S and filling in the other slots with the other cars. Then we would add up all of those permutations to get the total number of ways the cars can be arranged.

For instance we could start with PQS and then multiply by 4 x 3 x 2 x 1, the numbers in which would account for the number of choices available for each of the remaining slots. Then we could move S to the next slot and so on.

That's pretty time consuming though, and fortunately there is a much faster way.

If we just do all of the permutations, considering PQ as a unit with two versions, S will be to the right of P and Q half of the time and to the left of P and Q half of the time. So half of the permutations will fit the constraints, and our answer will be (2(6!))/2 = 6! = 720.

So the correct answer is B.

Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

Seven cars of seven different models are going to park in a row of seven side-by-side parking spots for an advertisement. Model P and Model Q must park next to each other, and Model S must be somewhere to the right of Models P & Q. How many possible configurations are there for the cars?

A.) 600
B.) 720
C.) 1440
D.) 4320
E.) 4800


Since Model P and Model Q must park next to each other we consider Model P and Model Q just as one Model. Then we have 6 Models.

Among 6 parking lots we choose two parking lot(among these two parking lots Model P and Model Q should be placed at left parking and Model S should be placed at right one. Then Model S is somewhere to the right of Models P & Q). The number of cases is 6C2 = 15.

Then among remaining 4 parking lots we distribute 4 remaining Models. The number of cases is 4!=24.

Moreover we must consider that Model P and Model Q can change their places.

So the total number is 15*24*2=720.

The answer is (B)

Seven cars of seven different models are going to park in a row of seven side-by-side parking spots for an advertisement. Model P and Model Q must park next to each other, and Model S must be somewhere to the right of Models P & Q. How many possible configurations are there for the cars?

A. 600
B. 720
C. 1440
D. 4320
E. 4800

Let the 7 cars be A, B, C, D, P, Q and S.

Since P and Q must occupy adjacent positions, consider PQ a single element in the arrangement.
The number of ways to arrange the 6 elements A, B, C, D, PQ and S = 6! = 720.

In 1/2 of these arrangements, S will be to the LEFT of PQ.
In the remaining 1/2 of these arrangements, S will be to the RIGHT of PQ.
Thus, the number of arrangements in which S is to the right of PQ = (1/2)(720).

Since PQ can switch to QP -- doubling the total number of possible arrangements -- we multiply by 2:
(2)(1/2)(720) = 720.

The correct answer is B.