1986 AJHSME Problems/Problem 9

Problem

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

$[asy] draw((0,0)--(3,0),MidArrow); draw((3,0)--(6,0),MidArrow); draw(6*dir(60)--3*dir(60),MidArrow); draw(3*dir(60)--(0,0),MidArrow); draw(3*dir(60)--(3,0),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--(6,0),MidArrow); draw(6*dir(60)--5.1961524227066318805823390245176*dir(30),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--3*dir(60),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--(3,0),MidArrow); label("M",6*dir(60),N); label("N",(6,0),SE); label("A",3*dir(60),NW); label("B",5.1961524227066318805823390245176*dir(30),NE); label("C",(3,0),S); label("D",(0,0),SW); [/asy]$

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

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 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 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 6 ways to get to N.

6 is E.

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1986 AJHSME Problems/Problem 9

Problem

Solution

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