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?
*This question is included in Sequencing: Lesson Set 4 (of 5) - Gaps, question #19
(A) | four |
(B) | ... |
(C) | ... |
(D) | ... |
(E) | ... |
(F) | ... |
The solution is
Posted: 01/21/2012 02:31
I can't understand the question. Can you explain pls.
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
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
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.
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.