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A class is composed 2 brothers and 6 other boys. In how many ways can all the boys be seated at the round table so that the 2 brothers are not seated besides each other?
- a)1440
- b)4320
- c)3600
- d)720
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A class is composed 2 brothers and 6 other boys. In how many ways can ...
Take 1 person from 6 and fix him and 5 others can arranged in -- 5! ways=120
there are 6 places left in which 2 brothers can sit
so they can choose any 2 places from 6 - 6C2 ways=15
2 brothers can arrange themselves in 2! ways=15*2=30
total ways=120*30=3600
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A class is composed 2 brothers and 6 other boys. In how many ways can ...
Solution:
Firstly, let's count the total number of ways in which all the boys can be seated at the round table. The number of ways of seating 8 boys at a round table is (8-1)! = 7!.
Now, let's consider the two brothers as a single entity. Therefore, we have 7 entities to arrange around the table.
Let's assume that the two brothers are sitting together. We can treat them as a single entity and arrange them with the remaining 6 boys. The number of ways of arranging the 7 entities in this case is 7! × 2! (since the two brothers can be arranged in 2 ways).
Now, let's count the number of arrangements in which the two brothers are not seated together. We can use the principle of complementary counting. Therefore, the number of arrangements in which the two brothers are not seated together is:
Total number of arrangements - Number of arrangements in which the two brothers are seated together
= 7! - 7! × 2!
= 7!(1 - 2/7)
= 6 × 5 × 4 × 3 × 2 × 1 × 5/7
= 3600
Therefore, the answer is option C (3600).
Firstly, let's count the total number of ways in which all the boys can be seated at the round table. The number of ways of seating 8 boys at a round table is (8-1)! = 7!.
Now, let's consider the two brothers as a single entity. Therefore, we have 7 entities to arrange around the table.
Let's assume that the two brothers are sitting together. We can treat them as a single entity and arrange them with the remaining 6 boys. The number of ways of arranging the 7 entities in this case is 7! × 2! (since the two brothers can be arranged in 2 ways).
Now, let's count the number of arrangements in which the two brothers are not seated together. We can use the principle of complementary counting. Therefore, the number of arrangements in which the two brothers are not seated together is:
Total number of arrangements - Number of arrangements in which the two brothers are seated together
= 7! - 7! × 2!
= 7!(1 - 2/7)
= 6 × 5 × 4 × 3 × 2 × 1 × 5/7
= 3600
Therefore, the answer is option C (3600).
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A class is composed 2 brothers and 6 other boys. In how many ways can all the boys be seated at the round table so that the 2 brothers are not seated besides each other?a)1440b)4320c)3600d)720Correct answer is option 'C'. Can you explain this answer?
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