There are 5 boys and 4 girls in my class.
In how many ways can they be seated in a row of 9 chairs such that at least 2 boys are next to each other?
In how many ways can they be seated in a row of 9 chairs such that at least 3 girls are all next to each other?
number of ways for them to sit without any restriction
= 9! = 362880
Let's calculate the number of ways for none of the boys to be together.
that can only happen this way
BGBGBGBGB
number of such ways
= 5x4x4x3x3x2x2x1 = 2880
So the number of ways with at least 2 boys together
= 362880-2880 = 360000
To find the number of ways the boys and girls can be seated in a row of 9 chairs such that at least 2 boys are next to each other, we can use the principle of complementary counting.
First, let's calculate the total number of ways they can be seated without any restrictions. This can be done using the permutation formula.
Total number of ways without restrictions = 9!
Next, let's find the number of ways where no two boys are seated together.
Since there are 5 boys, they need to be separated by at least one girl. We can think of the 5 boys as 5 blocks to be placed in the 9 chairs, with the girls filling the spaces between them.
The number of ways to arrange the 5 boys in 9 chairs without any restrictions is 9P5.
Now, let's consider the spaces between the boys. There are 5 boys, so we have 6 spaces between them where the girls can be placed. The number of ways to place the 4 girls in these spaces is 6P4.
Therefore, the number of ways to seat the boys and girls such that no two boys are seated together is 9P5 * 6P4.
Finally, we can find the number of ways where at least 2 boys are seated together by subtracting the number of ways where no two boys are seated together from the total number of ways without restrictions:
Number of ways where at least 2 boys are seated together = Total number of ways without restrictions - Number of ways where no two boys are seated together.
For the second question, we can follow a similar approach.
The total number of ways without any restrictions is still 9!.
To find the number of ways where at least 3 girls are seated together, we can first calculate the number of ways where no three girls are seated together.
Again, we can think of the 4 girls as 4 blocks to be placed in the 9 chairs, with the boys filling the spaces between them.
The number of ways to arrange the 4 girls in 9 chairs without restrictions is 9P4.
Now, let's consider the spaces between the girls. There are 4 girls, so we have 6 spaces between them where the boys can be placed. The number of ways to place the 5 boys in these spaces is 6P5.
Therefore, the number of ways to seat the boys and girls such that no three girls are seated together is 9P4 * 6P5.
Finally, we can find the number of ways where at least 3 girls are seated together by subtracting the number of ways where no three girls are seated together from the total number of ways without restrictions:
Number of ways where at least 3 girls are seated together = Total number of ways without restrictions - Number of ways where no three girls are seated together.