Number of Cases ③ Routes and Conditional Cases
This section covers problems involving shortest routes along grid paths and counting cases with conditions.
How to Count Routes
When you can move only right or up along grid paths, the number of routes to each intersection is the sum of the numbers of routes to the intersection on the left and the intersection below.
Example 1
How many shortest routes are there if you move 3 steps to the right and 2 steps up?
Solution:
The number of arrangements of 3 right moves (R) and 2 up moves (U)
5C2 ＝ (5×4)÷(2×1) ＝ 10 routes
How many shortest routes are there if you move 3 steps to the right and 2 steps up?
Solution:
The number of arrangements of 3 right moves (R) and 2 up moves (U)
5C2 ＝ (5×4)÷(2×1) ＝ 10 routes
Conditional Cases
Example 2
When arranging 5 people, A, B, C, D, and E, in a row, how many arrangements have A and B next to each other?
Solution:
Treat A and B as one group → arrangements of 4 elements: 4!＝24 arrangements
Arrangements within the A-and-B group: 2 arrangements
Total: 24×2＝48 arrangements
When arranging 5 people, A, B, C, D, and E, in a row, how many arrangements have A and B next to each other?
Solution:
Treat A and B as one group → arrangements of 4 elements: 4!＝24 arrangements
Arrangements within the A-and-B group: 2 arrangements
Total: 24×2＝48 arrangements
Key Points
- Routes: arrangements of m right moves and n up moves＝(m+n)Cm
- Condition of being next to each other → treat them as one group
- “Not〜” cases → total number of cases − number of cases satisfying the condition
Practice Problems
- How many shortest routes are there if you move 4 steps to the right and 3 steps up?
- When arranging A, B, C, D, and E in a row, how many arrangements do not have A and B next to each other?
Answers and Explanations
- Answer: 35 routes
Explanation: Arrangements of 4 right moves and 3 up moves. Choose 3 up moves from a total of 7 moves: 7C3＝（7×6×5）÷（3×2×1）＝210÷6＝35 routes. - Answer: 72 arrangements
Explanation: Total arrangements: 5!＝120 arrangements. Arrangements in which A and B are next to each other: 4!×2＝48 arrangements (treat A and B as one group, arrange the 4 elements, and consider the 2 possible orders within the AB group). Arrangements in which A and B are not next to each other: 120－48＝72 arrangements.