Discussion

If the Wooster arrives at some time before the Rockville, then exactly how many different orders are there in which the seven trains could arrive?
(A)four
(B)...
(C)...
(D)...
(E)...
(F)...
*This question is included in Sequencing: Lesson Set 4 (of 5) - Gaps, question #19

The solution is

Posted: 01/21/2012 02:31
I can't understand the question. Can you explain pls.
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Contributor
Posted: 01/21/2012 04:15
Tade,

Y or W = 4th
W < S < Y
R < ( T and V )
~( T, V )
~( V, T )
And in the question W < R will make
1-2-3-4-5-6-7
WSRYTQV
WSRYVQT
The sequence at the end TQV and VQT can be repeated after shifting R to the left, the rules allow that
WRSYTQV
WRSYVQT

That make A four the correct answer.

Niels
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Contributor
Posted: 01/30/2012 01:16
Tade,

This is a difficult game. Let's have a look at the Rules:

R1: 4th is Y -or- W
R2: W...S...Y
R3: R...T/V
R4: T & V non-consec
*Stem Rule: W...R

So the first thing we'll need to do is look at how the Stem Rule affects the other Rules.
When we combine Rule 2 and the Stem Rule, we get: W...R...T/V.

We also know that we have W...S...Y (Rule 2).

So since S, Y, R, T, and V must come after W, W cannot be 4th. Therefore (by Rule 1), Y is 4th.

Now we know that Y is 4th, so since S must come between W and Y, it must be either 2rd (if W is first) or 3rd (if W is second). But we know R must come after W (Stem Rule), so we know W cannot be second. Therefore, W must be first.

So, for the first 3 letters, we can either have W, R, S -or- W, S, R.

Now let's figure out what the last three letters will be. We know that Q is unrestricted, so let's try to use Q to sperate the T & V (since T and V cannot be consecutive, by Rule 4).

If we do that, we've got two possibilities for the last 3 letters: T, Q, V -or- V, Q, T.

Now we've got all the possible orders:
1. WSRYTQV
2. WSRYVQT
3. WRSYTQV
4. WRSYVQT

Choice (A) is our winner.

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