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

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~Solution by <math>BladeRunnerAUG</math> (Frank FYC)
 
~Solution by <math>BladeRunnerAUG</math> (Frank FYC)
 +
 +
==Solution 4 (Casework)==
 +
 +
Casework Solution:
 +
x-distribution: 1-1-1-1 (1 way to order)
 +
y-distribution: 1-1-1-1 (1 way to order)
 +
<math>\dbinom{8}{4} = 70</math> ways total
 +
x-distribution: 1-1-2 (3 ways to order)
 +
y-distribution: 1-1-2 (3 ways to order)
 +
<math>\dbinom{6}{3} \times 9= 180</math> ways total
 +
x-distribution: 1-1-1-1 (1 way to order)
 +
y-distribution: 1-1-2 (3 ways to order)
 +
<math>\dbinom{7}{3} \times 3= 105</math> ways total
 +
x-distribution: 1-1-2 (3 ways to order)
 +
y-distribution: 1-1-1-1 (1 way to order)
 +
<math>\dbinom{7}{3} \times 3= 105</math> ways total
 +
x-distribution: 1-1-1-1 (1 way to order)
 +
y-distribution: 2-2 (1 way to order)
 +
<math>\dbinom{6}{4} = 15</math> ways total
 +
x-distribution: 2-2 (1 way to order)
 +
y-distribution: 1-1-1-1 (1 way to order)
 +
<math>\dbinom{6}{4} = 15</math> ways total
 +
x-distribution: 1-1-2 (3 ways to order)
 +
y-distribution: 2-2 (1 way to order)
 +
<math>\dbinom{5}{3} \times 3 = 30</math> ways total
 +
x-distribution: 2-2 (1 way to order)
 +
y-distribution: 1-1-2 (3 ways to order)
 +
<math>\dbinom{5}{3} \times 3 = 30</math> ways total
 +
x-distribution: 2-2 (1 way to order)
 +
y-distribution: 2-2 (1 way to order)
 +
<math>\dbinom{4}{2} = 6</math> ways total
 +
<math>6+30+15+105+180+70+30+15+105=\boxed{556}</math>
 +
-fidgetboss_4000
  
 
==Video Solution==
 
==Video Solution==

Revision as of 10:39, 14 October 2019

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.

$(0,0): 1$

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

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

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

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

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

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

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

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

A diagram of the numbers:

5 - 20 - 71 - 207 - $\boxed{556}$

3 - 10 - 32 - 84 - 207

2 - 5 - 14 - 32 - 71

1 - 2 - 5 - 10 - 20

1 - 1 - 2 - 3 - 5

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.

Solution 3 (General Case)

Mark the total number of distinct sequences of jumps for the frog to reach the point $(x,y)$ as $\varphi (x,y)$. Consider for each point $(x,y)$ in the first quadrant, there are only $4$ possible points in the first quadrant for frog to reach point $(x,y)$, and these $4$ points are \[(x-1,y); (x-2,y); (x,y-1); (x,y-2)\]. As a result, the way to count $\varphi (x,y)$ is \[\varphi (x,y)=\varphi (x-1,y)+\varphi (x-2,y)+\varphi (x,y-1)+\varphi (x,y-2)\]

Also, for special cases, \[\varphi (0,y)=\varphi (0,y-1)+\varphi (0,y-2)\]

\[\varphi (x,0)=\varphi (x-1,0)+\varphi (x-2,0)\]

\[\varphi (x,1)=\varphi (x-1,1)+\varphi (x-2,1)+\varphi (x,0)\]

\[\varphi (1,y)=\varphi (1,y-1)+\varphi (1,y-2)+\varphi (0,y)\]

\[\varphi (1,1)=\varphi (1,0)+\varphi (0,1)\]

Start with $\varphi (0,0)=1$, use this method and draw the figure below, we can finally get \[\varphi (4,4)=556\] (In order to make the LaTeX thing more beautiful to look at, I put $0$ to make every number a $3$-digits number)

\[005-020-071-207-\boxed{556}\] \[003-010-032-084-207\] \[002-005-014-032-071\] \[001-002-005-010-020\] \[001-001-002-003-005\]

So the total number of distinct sequences of jumps for the frog to reach $(4,4)$ is $\boxed {556}$.

~Solution by $BladeRunnerAUG$ (Frank FYC)

Solution 4 (Casework)

Casework Solution: x-distribution: 1-1-1-1 (1 way to order) y-distribution: 1-1-1-1 (1 way to order) $\dbinom{8}{4} = 70$ ways total x-distribution: 1-1-2 (3 ways to order) y-distribution: 1-1-2 (3 ways to order) $\dbinom{6}{3} \times 9= 180$ ways total x-distribution: 1-1-1-1 (1 way to order) y-distribution: 1-1-2 (3 ways to order) $\dbinom{7}{3} \times 3= 105$ ways total x-distribution: 1-1-2 (3 ways to order) y-distribution: 1-1-1-1 (1 way to order) $\dbinom{7}{3} \times 3= 105$ ways total x-distribution: 1-1-1-1 (1 way to order) y-distribution: 2-2 (1 way to order) $\dbinom{6}{4} = 15$ ways total x-distribution: 2-2 (1 way to order) y-distribution: 1-1-1-1 (1 way to order) $\dbinom{6}{4} = 15$ ways total x-distribution: 1-1-2 (3 ways to order) y-distribution: 2-2 (1 way to order) $\dbinom{5}{3} \times 3 = 30$ ways total x-distribution: 2-2 (1 way to order) y-distribution: 1-1-2 (3 ways to order) $\dbinom{5}{3} \times 3 = 30$ ways total x-distribution: 2-2 (1 way to order) y-distribution: 2-2 (1 way to order) $\dbinom{4}{2} = 6$ ways total $6+30+15+105+180+70+30+15+105=\boxed{556}$ -fidgetboss_4000

Video Solution

On The Spot STEM : https://www.youtube.com/watch?v=v2fo3CaAhmM

2018 AIME II (Problems • Answer Key • Resources)
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