In how many ways can 5 boys and 5 girls be seated around a table so that no 2 boys are adjacent?
Clearly they must alternate boy-girl-boy-girl etc. So the question becomes: How many ways can that be done?
Consider chair #1. If a boy goes there, there are 5 possibilities. Then there are 5 for the next chair (any girl), then any of four boys, etc until you get a number of permutations of 5!*5! = 14,400
If chair #1 starts with a girl, then there are 14,400 more possible permutations. That is a total of 28,800
If you are only interested in the sequence of people and not who sits on chair #1, then you must divide by 10, giving 2880.
Lets arrange the 5 girls first, fixing 1: (5-1)! = 4! = 24 .
Then arrange 5 boys = 5! = 120
so, total arrangements required: 120 x 24 = 2880
To find the number of ways the boys and girls can be seated around the table such that no two boys are adjacent, we need to consider the positions of the boys and girls separately.
Step 1: Find the number of ways to arrange the boys without any restrictions.
Since there are 5 boys, they can be seated around the table in 5! (5 factorial) ways.
Step 2: Find the number of ways to arrange the girls without any restrictions.
Similarly, since there are 5 girls, they can be seated around the table in 5! ways.
Step 3: Find the number of ways to arrange the boys and girls together such that no two boys are adjacent.
In order for no two boys to be adjacent, we need to alternate the seating arrangement of boys and girls.
We can start by placing a boy at any of the 5 seats. There are 5 possible choices for the first boy's seat.
Once the first boy is seated, we can place the remaining boys in the alternate seats (every second seat). There are 4 boys left to be seated, and 4 alternate seats available.
After the boys are seated, the remaining seats are filled with the girls in 5! ways.
Therefore, the total number of ways to seat the boys and girls around the table such that no two boys are adjacent is:
5! x 4! x 5!
= 120 x 24 x 120
= 345,600
So, there are 345,600 ways to seat 5 boys and 5 girls around a table such that no two boys are adjacent.
To solve this problem, we can use the concept of permutations.
Step 1: Calculate the total number of ways to seat the 10 people around the table without any restrictions. This can be done using the formula for circular permutations, which is (n-1)!, where n is the number of people to be seated. In this case, we have 10 people, so the total number of ways to seat them around the table is (10-1)! = 9!.
Step 2: Calculate the number of ways to seat the boys in a way that no 2 boys are adjacent. We can treat the 5 boys as 5 distinct objects. The number of ways to arrange these 5 boys in a row is 5!. However, since this is a circular arrangement, we need to divide the result by 5 to account for the different starting positions that yield the same circular arrangement. Therefore, the number of ways to seat the boys without any restrictions is 5! / 5 = 4!.
Step 3: Calculate the number of ways to seat the boys and girls together without any restrictions. Since the girls can be seated in any order, we can use the same approach as in Step 2. The number of ways to arrange the 5 girls in a row is 5!. Dividing this by 5 gives us the number of circular arrangements, which is 4!.
Step 4: Calculate the number of ways to seat the boys and girls with the restriction that no 2 boys are adjacent. Taking into account the seating arrangement for the boys from Step 2, we need to find the number of ways to arrange the 5 girls among the 4 empty spaces between the boys. This can be calculated using the concept of combinations. We can choose 5 positions out of the 4 empty spaces to place the girls. The number of combinations is given by 4C5 = 4! / [(4-5)! * 5!] = 4! = 4.
Step 5: Calculate the final answer by subtracting the result from Step 4 from the result from Step 1: 9! - (4! * 4).
Therefore, there are 9! - (4! * 4) = 326,880 - (24 * 4) = 326,880 - 96 = 326,784 ways to seat 5 boys and 5 girls around a table such that no 2 boys are adjacent.