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1986 AJHSME Problems/Problem 9

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Problem

Using only the paths and the directions shown, how many different routes are there from \text{M} to \text{N}?

draw((0,0)--(3,0),MidArrow);draw((3,0)--(6,0),MidArrow);draw(6*dir(60)--3*dir(60),MidArrow);draw(3*dir(60)--(0,0),MidArrow);d...

\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6

Solution

There is 1 way to get from C to N. There is only one way to get from D to N, which is DCN.

Since A can only go to C or D, which each only have 1 way to get to N each, there are 1+1=2 ways to get from A to N.

Since B can only go to A, C or N, and A only has 2 ways to get to N, C only has 1 way and to get from B to N is only 1 way, there are 2+1+1=4 ways to get from B to N.

M can only go to either B or A, A has 2 ways and B has 4 ways, so M has 4+2=6 ways to get to N.

6 is \boxed{\text{E}}.

A diagram labeled with the number of ways to get to \text{N} from each point might look like

draw((0,0)--(3,0),MidArrow);draw((3,0)--(6,0),MidArrow);draw(6*dir(60)--3*dir(60),MidArrow);draw(3*dir(60)--(0,0),MidArrow);d...

See Also

1986 AJHSME (ProblemsResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
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