CBSEGrade 11MathematicsPermutations and Combinations

7 Friends and 7 Chairs

In a party, 7 friends are to be seated on 7 different chairs. However, two of the friends insist on sitting together in the first two chairs. How many ways are there to arrange the friends for the seating arrangement, considering the restriction?

💬 1 answers0 votes👁 9 views29 June 2026

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📌 CONCEPT: The problem involves finding the number of ways to arrange the friends considering a restriction that two friends must sit together in the first two chairs.

📐 RULE / FORMULA: We can use the concept of permutations for this problem. The total number of ways to arrange the friends is given by the product of the permutations of the two 'sitting-together' friends and the remaining 5 friends, and the permutations of the 7 chairs.

💡 WORKED EXAMPLE: Consider the two friends who insist on sitting together as a single entity. There are 6! ways to arrange the remaining 6 entities (5 friends and 1 'sitting-together' entity) in 6 chairs. Within this entity, there are 2! ways to arrange the two friends. Therefore, the total number of ways to arrange the friends is 6! * 2!.

⚠️ COMMON MISTAKE: Students often forget to consider the restriction of the two friends sitting together and simply calculate the total number of permutations of 7 friends and 7 chairs, which is 7!. However, this does not take into account the restriction of the two friends sitting together.

29 Jun 26

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Permutations and Combinations

Mathematics · Grade 11

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