Difference between revisions of "2018 AMC 10B Problems/Problem 18"
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<math>A_1 \circ \circ \ \circ \ \circ \ \circ </math> | <math>A_1 \circ \circ \ \circ \ \circ \ \circ </math> | ||
− | Then there are 4 choices for the second seat (<math>B_1</math>, <math>B_2</math>, <math>C_1</math>, or <math>C_2</math>). | + | Then there are 4 choices for the second seat (<math>B_1</math>, <math>B_2</math>, <math>C_1</math>, or <math>C_2</math>). Like before, it doesn't matter who takes the seat, so WLOG suppose it is <math>B_1</math>: |
<math>A_1 B_1 \circ \ \circ \ \circ \ \circ</math> | <math>A_1 B_1 \circ \ \circ \ \circ \ \circ</math> | ||
− | The last seat in the first row cannot be <math>A_2</math> because it would be impossible to create a second row that satisfies the conditions. Therefore, it must be <math>C_1</math> or <math>C_2</math>. | + | The last seat in the first row cannot be <math>A_2</math> because it would be impossible to create a second row that satisfies the conditions. Therefore, it must be <math>C_1</math> or <math>C_2</math>. Let's say WLOG that it is <math>C_1</math>. There are two ways to create a second row:,,,,' |
<math>A_1 B_1 C_1 \ B_2 C_2 A_2</math> | <math>A_1 B_1 C_1 \ B_2 C_2 A_2</math> | ||
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Therefore, there are <math>6 \cdot 4 \cdot 2 \cdot 2= \boxed{\textbf{(D)} \text{ 96}}</math> possible seating arrangements. | Therefore, there are <math>6 \cdot 4 \cdot 2 \cdot 2= \boxed{\textbf{(D)} \text{ 96}}</math> possible seating arrangements. | ||
− | Written by: | + | Written by: HoloGuard1728 |
==See Also== | ==See Also== |
Revision as of 18:46, 20 December 2018
Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?
Solution 1 (Casework)
We can begin to put this into cases. Let's call the pairs , and , and assume that a member of pair is sitting in the leftmost seat of the second row. We can have the following cases then.
Case 1: Second Row: a b c Third Row: b c a
Case 2: Second Row: a c b Third Row: c b a
Case 3: Second Row: a b c Third Row: c a b
Case 4: Second Row: a c b Third Row: b a c
For each of the four cases, we can flip the siblings, as they are distinct. So, each of the cases has possibilities. Since there are four cases, when pair has someone in the leftmost seat of the second row, there are 32 ways to rearrange it. However, someone from either pair , , or could be sitting in the leftmost seat of the second row. So, we have to multiply it by 3 to get our answer of . So, the correct answer is .
Written By: Archimedes15
Solution 2
Call the siblings , , , , , and .
There are 6 choices for the child in the first seat, and it doesn't matter which one takes it, so suppose Without loss of generality that takes it (a is an empty seat):
Then there are 4 choices for the second seat (, , , or ). Like before, it doesn't matter who takes the seat, so WLOG suppose it is :
The last seat in the first row cannot be because it would be impossible to create a second row that satisfies the conditions. Therefore, it must be or . Let's say WLOG that it is . There are two ways to create a second row:,,,,'
Therefore, there are possible seating arrangements.
Written by: HoloGuard1728
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 AMC 10 Problems and Solutions |
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