The diagram above shows the various paths along which a mous

First recognize that, in order to get from point X to point Y, we MUST travel through points A,B,C,D,E and F.

So, we can take the task of getting from point X to Y and break it into stages.

Stage 1: Move from point X to point A
There's only 1 possible route, so we can complete stage 1 in 1 way.

Stage 2: Move from point A to point B
There are 2 possible routes, so we can complete stage 2 in 2 ways.

Stage 3: Move from point B to point C
There's only 1 possible route, so we can complete stage 3 in 1 way.

Stage 4: Move from point C to point D
There are 2 possible routes, so we can complete stage 4 in 2 ways.

Stage 5: Move from point D to point E
There's only 1 possible route, so we can complete stage 5 in 1 way.

Stage 6: Move from point E to point F
There are 3 possible routes, so we can complete stage 6 in 3 ways.

Stage 7: Move from point F to point Y
There's only 1 possible route, so we can complete stage 7 in 1 way.

By the Fundamental Counting Principle (FCP), we can complete all 7 stages (and thus move from point X to point Y) in (1)(2)(1)(2)(1)(3)(1) ways (= 12 ways)

Answer: C

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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Also counting is fast

12 or 11 paths. 12 is the only among the options.

So C

Dear Bunnel,
Could you please clarify it more...How the forks are working?

is pretty simple: one the first fork you have 2 choices - right and left; idem for the second one; 3 for the third one: right, left and central to the goal. 2*2*3=12

That's it

Technique here is to multiply the number of choices in every point of decision:
\(2*2*3 = 12\)

Answer: C

Hi Brunel - Would you help us with more such questions?

Why is it multiplied here ? Why can't we add all options ?

Posted from my mobile device

Because of Principle of Multiplication: if one event can occur in m ways and a second can occur independently of the first in n ways, then the two events can occur in m*n ways.

For example, if you have two pairs of shoes, A and B, and two shirts, X and Y, then there will be 2*2 = 4 shoes-shirt combinations:
AX;
AY;
BX;
BY.

Hope it's clear.

The best way to solve this problem is to use the idea of the fundamental counting principle. In a more standard form you could be asked a question, such as if Tom as 3 belts, 4 ties, and 6 shirts, how many outfits could he make with those items? We can consider each item a decision point, i.e., belts, ties, and shirts. To solve this, we just need to multiply the number of decisions Tom can make together, so:

3 x 4 x 6 = 72 ways.

Tom has 72 options when dressing with those items.

This same logic can be applied to this problem here. We can first determine the number ways the mouse can go from one point to the next.

X to A = 1

A to B = 2

B to C = 1

C to D= 2

D to E = 1

E to F = 3

F to Y =1

Therefore, to find the total number of ways from X to Y we can multiply all these numbers together:

1 x 2 x 1 x 2 x 1 x 3 x 1 = 12 ways.

There are 12 different paths.

Answer is C.

can anyone answer this question using combinatorics please?

For the first point you have two options (2C1) for the C, same for the second point (2C1), for third point you have 3 options (3C1).

--> 2C1 * 2C1 * 3C1 = 2*2*3 = 12

We will do addition here also to answer the question posted earlier.
a We will first add 2 choice[1+1]
b Then we will add 2 choices [1+1]
c Then we will add 3 choices [1+1+1]
Then we will multiply the straight lines first to get 1
Then we will multiply the different probabilities a,b,c to get 12

Here we multiply because all tasks are required i.e. we can't go directly from x to y.

DIFFERENT ways to go from x to y= 2*2*3= 12

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