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The total number of ways in which six + signs and 4 - 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 + signs and 4 - signs can be arr...
Understanding the Problem
To find the number of ways to arrange six + signs and four - signs such that no two - signs are adjacent, we can follow a systematic approach.
Arranging the + Signs
1. Start by arranging the + signs:
We first arrange the six + signs in a line.
This creates gaps where the - signs can be placed.
2. Identifying Gaps:
When the six + signs are arranged, the gaps can be visualized as follows:
+ + + + + +
The gaps available for placing - signs are:
- Before the first +
- Between + signs (5 gaps)
- After the last +
This gives us a total of 7 gaps.
Placing the - Signs
1. Placing - Signs in the Gaps:
To ensure that no two - signs are adjacent, we can place one - sign in each of the selected gaps.
We need to select 4 gaps from the 7 available.
2. Calculating Combinations:
The number of ways to choose 4 gaps from 7 can be calculated using combinations:
This is given by the formula C(7, 4) = 7! / (4! * (7 - 4)!) = 35.
Final Count of Arrangements
Therefore, the total number of arrangements of six + signs and four - signs such that no two - signs occur together is 35.
To find the number of ways to arrange six + signs and four - signs such that no two - signs are adjacent, we can follow a systematic approach.
Arranging the + Signs
1. Start by arranging the + signs:
We first arrange the six + signs in a line.
This creates gaps where the - signs can be placed.
2. Identifying Gaps:
When the six + signs are arranged, the gaps can be visualized as follows:
+ + + + + +
The gaps available for placing - signs are:
- Before the first +
- Between + signs (5 gaps)
- After the last +
This gives us a total of 7 gaps.
Placing the - Signs
1. Placing - Signs in the Gaps:
To ensure that no two - signs are adjacent, we can place one - sign in each of the selected gaps.
We need to select 4 gaps from the 7 available.
2. Calculating Combinations:
The number of ways to choose 4 gaps from 7 can be calculated using combinations:
This is given by the formula C(7, 4) = 7! / (4! * (7 - 4)!) = 35.
Final Count of Arrangements
Therefore, the total number of arrangements of six + signs and four - signs such that no two - signs occur together is 35.
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The total number of ways in which six + signs and 4 - 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 + signs and 4 - 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 + signs and 4 - 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 + signs and 4 - signs can be arranged in a line such that no two - signs occur together is?.
The total number of ways in which six + signs and 4 - 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 + signs and 4 - 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 + signs and 4 - signs can be arranged in a line such that no two - signs occur together is?.
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