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The total number of ways in which six ' ' and four '-' signs can be arranged in a line such that no two '-' signs occur together is?
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The total number of ways in which six ' ' and four '-' signs can be a...
Understanding the Problem
We need to arrange six ' ' (spaces) and four '-' (dashes) such that no two dashes occur together.
Step 1: Arranging the Spaces
- First, we arrange the six spaces. This creates potential slots for the dashes.
- The arrangement of six spaces can be visualized as:
S S S S S S
- This arrangement creates seven potential slots for placing dashes:
_ S _ S _ S _ S _ S _ S _
Step 2: Placing the Dashes
- We have seven slots (represented by underscores) in which we can place our four dashes.
- To ensure that no two dashes are adjacent, we need to select four out of these seven slots.
Step 3: Choosing the Slots
- The number of ways to select four slots from seven can be calculated using combinations.
- We use the combination formula:
C(n, r) = n! / (r!(n-r)!)
- Here, n = 7 (total slots) and r = 4 (dashes to place).
Step 4: Calculation
- C(7, 4) = 7! / (4! * (7-4)!)
- This simplifies to:
C(7, 4) = 7! / (4! * 3!) = (7 * 6 * 5) / (3 * 2 * 1) = 35
Conclusion
- Therefore, the total number of ways to arrange six spaces and four dashes such that no two dashes are adjacent is 35.
We need to arrange six ' ' (spaces) and four '-' (dashes) such that no two dashes occur together.
Step 1: Arranging the Spaces
- First, we arrange the six spaces. This creates potential slots for the dashes.
- The arrangement of six spaces can be visualized as:
S S S S S S
- This arrangement creates seven potential slots for placing dashes:
_ S _ S _ S _ S _ S _ S _
Step 2: Placing the Dashes
- We have seven slots (represented by underscores) in which we can place our four dashes.
- To ensure that no two dashes are adjacent, we need to select four out of these seven slots.
Step 3: Choosing the Slots
- The number of ways to select four slots from seven can be calculated using combinations.
- We use the combination formula:
C(n, r) = n! / (r!(n-r)!)
- Here, n = 7 (total slots) and r = 4 (dashes to place).
Step 4: Calculation
- C(7, 4) = 7! / (4! * (7-4)!)
- This simplifies to:
C(7, 4) = 7! / (4! * 3!) = (7 * 6 * 5) / (3 * 2 * 1) = 35
Conclusion
- Therefore, the total number of ways to arrange six spaces and four dashes such that no two dashes are adjacent is 35.
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The total number of ways in which six ' ' and four '-' signs can be arranged in a line such that no two '-' signs occur together is?
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The total number of ways in which six ' ' and four '-' signs can be arranged in a line such that no two '-' signs occur together is? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about The total number of ways in which six ' ' and four '-' signs can be arranged in a line such that no two '-' signs occur together is? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The total number of ways in which six ' ' and four '-' signs can be arranged in a line such that no two '-' signs occur together is?.
The total number of ways in which six ' ' and four '-' signs can be arranged in a line such that no two '-' signs occur together is? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about The total number of ways in which six ' ' and four '-' signs can be arranged in a line such that no two '-' signs occur together is? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The total number of ways in which six ' ' and four '-' signs can be arranged in a line such that no two '-' signs occur together is?.
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