Compute the permutations and combinations.
in how many ways can 5 boys and 5 girls be seated at a round table so that the boys and girls are seated alternately?

Name the chairs be chair 1, 2, 3..., 9, 10.

Assume the chairs are on a line, not on a circle.

Let one of the boys be Adam and let Adam sit on chair 1.

the seats 3, 5, 7 and 9 are to be occupied by boys, and the seats 2, 4, 6, 8 and 10 by girls:



This can be done by (4*3*2*1) * (5*4*3*2*1) that is 4!*5! ways.

Now back in the round table.

Consider one particular arrangements of the boys and girls , where Adam is in the first seat.

Take each of the boys and girls to the next chair on their right.

On the linear row, Adam now is seated in the chair 2, but since everyone has moved 1 chair, it is not a different arrangement, in the round table.

In this way, by moving Adam together with his friends in any of the 10 chairs, without changing their relative position, we see that 

4!*5! is made up of different arrangements, each of which is counted 10 times.

so,  there are (4!5!)/10= 4*3 * 4!=288 different arrangements in a round table.

288
 

X^2 - 4x +(4/2^2) + y^2 + 8y + (8/2^2) = -4 + (4/2^2) + (8/2^2)(x + 2)^2 + (y + 4)^2 = -4 + 4 + 16your center is (-2, -4) and the radius is the sqrt of -4 + 4 + 16 which is 4