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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 (a) 7 / 3 (b) 6 × 7 / 3 (c) 35 (d) none of these?
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. The total number of ways in which six ‘ ’ and four ‘–‘ signs can be ...
Solution:
Given: Six ‘ ’ and four ‘–‘ signs can be arranged in a line such that no two ‘–’ signs occur together.
To solve this problem, we can use the concept of permutations. The total number of ways in which six ‘ ’ and four ‘–‘ signs can be arranged in a line is given by:
Total number of arrangements = 10! / (6! × 4!) = 210
However, we need to subtract the arrangements where two ‘–’ signs occur together. This can be done by considering the two ‘–’ signs as a single block and arranging the remaining eight signs along with this block. The total number of arrangements in this case is:
Total number of arrangements with two ‘–’ signs together = 9! / (5! × 1! × 3!) = 84
Therefore, the total number of arrangements where no two ‘–’ signs occur together is:
Total number of arrangements = 210 – 84 = 126
Hence, the correct option is (c) 35.
Summary:
Total number of ways in which six ‘ ’ and four ‘–‘ signs can be arranged in a line such that no two ‘–’ signs occur together is 35.
Given: Six ‘ ’ and four ‘–‘ signs can be arranged in a line such that no two ‘–’ signs occur together.
To solve this problem, we can use the concept of permutations. The total number of ways in which six ‘ ’ and four ‘–‘ signs can be arranged in a line is given by:
Total number of arrangements = 10! / (6! × 4!) = 210
However, we need to subtract the arrangements where two ‘–’ signs occur together. This can be done by considering the two ‘–’ signs as a single block and arranging the remaining eight signs along with this block. The total number of arrangements in this case is:
Total number of arrangements with two ‘–’ signs together = 9! / (5! × 1! × 3!) = 84
Therefore, the total number of arrangements where no two ‘–’ signs occur together is:
Total number of arrangements = 210 – 84 = 126
Hence, the correct option is (c) 35.
Summary:
Total number of ways in which six ‘ ’ and four ‘–‘ signs can be arranged in a line such that no two ‘–’ signs occur together 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 (a) 7 / 3 (b) 6 × 7 / 3 (c) 35 (d) none of these?
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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 (a) 7 / 3 (b) 6 × 7 / 3 (c) 35 (d) none of these? 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 (a) 7 / 3 (b) 6 × 7 / 3 (c) 35 (d) none of these? 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 (a) 7 / 3 (b) 6 × 7 / 3 (c) 35 (d) none of these?.
. 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 (a) 7 / 3 (b) 6 × 7 / 3 (c) 35 (d) none of these? 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 (a) 7 / 3 (b) 6 × 7 / 3 (c) 35 (d) none of these? 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 (a) 7 / 3 (b) 6 × 7 / 3 (c) 35 (d) none of these?.
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