AMC10 2008 B

Answer (C): Let the women be seated first. The first woman may sit in any of the 10 chairs. Because men and women must alternate, the number of choices for the remaining women is 4, 3, 2, and 1. Thus the number of possible seating arrangements for the women is $10\cdot 4! = 240$. Without loss of generality, suppose that a woman sits in chair 1. Then this woman's spouse must sit in chair 4 or chair 8. If he sits in chair 4 then the women sitting in chairs 7, 3, 9, and 5 must have their spouses sitting in chairs 10, 6, 2, and 8, respectively. If he sits in chair 8 then the women sitting in chairs 5, 9, 3, and 7 must have their spouses sitting in chairs 2, 6, 10, and 4, respectively. So for each possible seating arrangement for the women there are two arrangements for the men. Hence, there are $2\cdot 240 = 480$ possible seating arrangements.