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Difference between revisions of "1986 AJHSME Problems/Problem 9"
 
(2 intermediate revisions by one other user not shown)
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==Solution==
 
==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.
+
There is 1 way to get from C to N. There is only one way to get from D to N, which is DCN.  
  
−
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.
+
Since A can only go to C or D, which each only have 1 way to get to N each, there are <math>1+1=2</math> ways to get from A to N.  
  
−
6 is E.
+
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 <math>2+1+1=4</math> 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 <math>4+2=6</math> ways to get to N.
 +
 
 +
6 is <math>\boxed{\text{E}}</math>.
 +
 
 +
A diagram labeled with the number of ways to get to <math>\text{N}</math> from each point might look like
 +
 
 +
<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);
 +
label("$1$",(6,0),NW,blue);
 +
label("$1$",(3,.1),N,blue);
 +
label("$1$",(.2,0),NE,blue);
 +
label("$2$",3*dir(60),SE,blue);
 +
label("$4$",5.1961524227066318805823390245176*dir(30),SW,blue);
 +
label("$6$",6*dir(60),S,blue);
 +
</asy>
  
 
==See Also==
 
==See Also==
  
−
[[1986 AJHSME Problems]]
+
{{AJHSME box|year=1986|num-b=8|num-a=10}}
 +
[[Category:Introductory Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 20:13, 3 July 2013

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 $1+1=2$ 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

[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); label("$1$",(6,0),NW,blue); label("$1$",(3,.1),N,blue); label("$1$",(.2,0),NE,blue); label("$2$",3*dir(60),SE,blue); label("$4$",5.1961524227066318805823390245176*dir(30),SW,blue); label("$6$",6*dir(60),S,blue); [/asy]

See Also

1986 AJHSME (Problems • Answer Key • Resources)
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
All AJHSME/AMC 8 Problems and Solutions

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