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Difference between revisions of "2018 AIME II Problems/Problem 8"

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<math>(0,0): 1</math>
 
<math>(0,0): 1</math>
  
−
<math>(1,0)=(0,1)=
+
<math>(1,0)=(0,1)=1</math>
  
−
</math>(2,0)=(0, 2)=2<math>
+
<math>(2,0)=(0, 2)=2</math>
  
−
</math>(3,0)=(0, 3)=3<math>
+
<math>(3,0)=(0, 3)=3</math>
  
−
</math>(4,0)=(0, 4)=5<math>
+
<math>(4,0)=(0, 4)=5</math>
  
−
</math>(1,1)=2<math>, </math>(1,2)=(2,1)=5<math>, </math>(1,3)=(3,1)=10<math>, </math>(1,4)=(4,1)= 20<math>
+
<math>(1,1)=2</math>, <math>(1,2)=(2,1)=5</math>, <math>(1,3)=(3,1)=10</math>, <math>(1,4)=(4,1)= 20</math>
  
−
</math>(2,2)=14, (2,3)=(3,2)=32, (2,4)=(4,2)=71<math>
+
<math>(2,2)=14, (2,3)=(3,2)=32, (2,4)=(4,2)=71</math>
  
−
</math>(3,3)=84, (3,4)=(4,3)=207<math>
+
<math>(3,3)=84, (3,4)=(4,3)=207</math>
  
−
</math>(4,4)=2\cdot \left( (4,2)+(4,3)\right) = 2\cdot \left( 207+71\right)=2\cdot 278=\boxed{556}$
+
<math>(4,4)=2\cdot \left( (4,2)+(4,3)\right) = 2\cdot \left( 207+71\right)=2\cdot 278=\boxed{556}</math>
  
 
==Solution 2==
 
==Solution 2==

Revision as of 06:38, 27 March 2018

Problem

A frog is positioned at the origin of the coordinate plane. From the point $(x, y)$, the frog can jump to any of the points $(x + 1, y)$, $(x + 2, y)$, $(x, y + 1)$, or $(x, y + 2)$. Find the number of distinct sequences of jumps in which the frog begins at $(0, 0)$ and ends at $(4, 4)$.

Solution 1

We solve this problem by working backwards. Notice, the only points the frog can be on to jump to $(4,4)$ in one move are $(2,4),(3,4),(4,2),$ and $(4,3)$. This applies to any other point, thus we can work our way from $(0,0)$ to $(4,4)$, recording down the number of ways to get to each point recursively.

$(0,0): 1$

$(1,0)=(0,1)=1$

$(2,0)=(0, 2)=2$

$(3,0)=(0, 3)=3$

$(4,0)=(0, 4)=5$

$(1,1)=2$, $(1,2)=(2,1)=5$, $(1,3)=(3,1)=10$, $(1,4)=(4,1)= 20$

$(2,2)=14, (2,3)=(3,2)=32, (2,4)=(4,2)=71$

$(3,3)=84, (3,4)=(4,3)=207$

$(4,4)=2\cdot \left( (4,2)+(4,3)\right) = 2\cdot \left( 207+71\right)=2\cdot 278=\boxed{556}$

Solution 2

We'll refer to the moves $(x + 1, y)$, $(x + 2, y)$, $(x, y + 1)$, and $(x, y + 2)$ as $R_1$, $R_2$, $U_1$, and $U_2$, respectively. Then the possible sequences of moves that will take the frog from $(0,0)$ to $(4,4)$ are all the permutations of $U_1U_1U_1U_1R_1R_1R_1R_1$, $U_2U_1U_1R_1R_1R_1R_1$, $U_1U_1U_1U_1R_2R_1R_1$, $U_2U_1U_1R_2R_1R_1$, $U_2U_2R_1R_1R_1R_1$, $U_1U_1U_1U_1R_2R_2$, $U_2U_2R_2R_1R_1$, $U_2U_1U_1R_2R_2$, and $U_2U_2R_2R_2$. We can reduce the number of cases using symmetry.

Case 1: $U_1U_1U_1U_1R_1R_1R_1R_1$

There are $\frac{8!}{4!4!} = 70$ possibilities for this case.

Case 2: $U_2U_1U_1R_1R_1R_1R_1$ or $U_1U_1U_1U_1R_2R_1R_1$

There are $2 \cdot \frac{7!}{4!2!} = 210$ possibilities for this case.

Case 3: $U_2U_1U_1R_2R_1R_1$

There are $\frac{6!}{2!2!} = 180$ possibilities for this case.

Case 4: $U_2U_2R_1R_1R_1R_1$ or $U_1U_1U_1U_1R_2R_2$

There are $2 \cdot \frac{6!}{2!4!} = 30$ possibilities for this case.

Case 5: $U_2U_2R_2R_1R_1$ or $U_2U_1U_1R_2R_2$

There are $2 \cdot \frac{5!}{2!2!} = 60$ possibilities for this case.

Case 6: $U_2U_2R_2R_2$

There are $\frac{4!}{2!2!} = 6$ possibilities for this case.

Adding up all these cases gives us $70+210+180+30+60+6=\boxed{556}$ ways.

2018 AIME II (Problems • Answer Key • Resources)
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