How many ways are there to go from A to D, passing through

Question

How many ways are there to go from A to D, passing through both B and C? You can only go north or east?
(See the attached) 

I can't even solve this simple work.. please anybody kind for the explanation....

KarishmaB · Accepted Answer

First of all, the question is definitely not 'this simple'. Though once you know how to deal with such questions, it becomes quite easy.

I do agree with vicksikand. Its easy to solve it his way. Let me elaborate on the theory behind it.

When I want to go from A to B, I have to take 1 step north and 1 step east. I can do this in two ways: First north, then east or first east, then north. I can say that I have two steps N and E and I have to arrange them. I can do it in 2 ways (NE) or (EN).

When I want to go from B to C, I have to take 1 step north and 2 steps east. I can do it in three ways: First north, then east, then east (NEE) or First east, then north, then east (ENE) or first east, then east, then north (EEN).
I can say I have 3 steps NEE and I have to arrange these in different ways. It can be done in 3!/2! ways = 3 ways (We divide by 2! because we have 2 E's. For more details check permutations theory)

When I want to go from C to D, we need to take two steps north and two east. That is, we have to arrange NNEE is different ways. This can be done in 4!/2!*2! = 6 ways (NNEE) or (NENE) or (NEEN) or (EENN) or (ENEN) or (ENNE)
We divide by two 2! because N is twice and E is twice.

Total number of ways of going from A to D = 2*3*6 = 36

vicksikand · Answer

A more structured way of looking at it:
A-B: 1 U(up) and 1 R(right), number of ways = 2 !/(1!*1!) = 2
B-C: 1 U and 2 R, number of ways= (1+2)!/(1!*2!)=3!/2! = 3
C-D: 2 U and 2 R, " = (2+2)!/(2!*2!) = 6
Total number of ways: 2*3*6 = 36

In such problems break the question into the number of U / D / L / R 's(up/down/left/right)
In almost all such problems it is given that the flow / movement is unidirectional (you cant move up in one step and down in the next).

For each such scenario determine the number of ways: (U + R)! / (U! * R!) :U, R for this example
We divide by U!(or R!) - since each U(or R) is identical.

On a similar note - 
find the number of different words you can formulate from 
DADDY = 5! / (3!)

GEORGE = 6! / (2!*2!)

sportyrizwan · Answer

Is the answer 2*3*6 = 36

For reaching from A to B there are 2 ways

For reaching from B to C there are 3 ways

For reaching from C to D there are 6 ways

Therefore there are total 36 different ways to reach from A to D

:lol:

pretttyune · Answer

the sily questions,, but I really thank you guys for your kind, persistent answers....

srivas · Answer

Yes. The total number of ways is 2 * 3 * 6 = 36

ankithakhu · Answer

There are actually four ways to go from B to C. so total number of ways = 2*4*6 = 48

niraj92 · Answer

From B to C we can go in 4 ways.
3h and 1v 4!/3!=4
Bunuel, could u please check the answer.

Bunuel · Answer

36 is correct: 2 ways from A to B, 3 ways from B to C (2h and 1v: 3!/2!=3) and 6 ways from C to D --> 2*3*6=36.

JJ2014 · Answer

Is this a typical GMAT problem?? I've never seen problems like this. Is this a PS problem?

Bunuel · Answer

Similar (though easier) question from OG: the-diagram-above-shows-the-various-paths-along-which-a-mous-144271.html

bumpbot · Answer

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How many ways are there to go from A to D, passing through

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How many ways are there to go from A to D, passing through

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