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