Arrangement in a row?

Question

In a mass wedding ceremony, 15 married couples are to be seated in 3 rows with 10 seats in each row. Find the number of ways of seating them in each of the following two cases:
(a) the partners of each couple sit next to each other
(b) the partners of each couple sit next to each other and, in addition, no person sit next to another person of the same gender

I am grateful for any help. Thanks!

Answer 1

(a)
For the seating configurations to be counted out, the possibilites for each seat must be multiplied together.
First seat 30 possibilities, Second seat 1 possbility (the spouse)
Third seat 28 possbilities; Fourth seat 1 possibility (the spouse)
so the total number of configurations are:
30 x 28 x 26 x... x 2 = 4.28499E+16

(b)
Similar to the solution above, the seating configurations must be counted.
First Seat 30 possibilities; second seat 1 (the spouse)
Third Seat 14 possibilities (only the different gender); fourth seat 1 (the spouse)
This continues till the tenth seat (end of first row)

For second row first seat 20 possibilities (all guests not seated)
second seat 1 (the spouse);
third seat 8 possibilities (only different gender guests); fourth seat 1 (the spouse).
This continues till the tenth seat on the second row (end of second row)

For third row first seat 10 possibilites (all guests not seated)
second seat 1 (the spouse)
third seat 4 possbilitied (only different gender guests); fourth seat 1 (the spouse)

number of configurations:
30 x 14 x 13 x 12 x 11
x 20 x 9 x 8 x 7 x 6
x 10 x 4 x 3 x 2
= 1.04614E+13

Answer 2

It makes no difference if it is 3 rows of 10 or a single row of 30.
(a)
1st 2 seats 15 couples times 2 ways of seating a couple = 2*15
2nd 2 seats 14 couples times 2 ways of seating a couple = 2*14==>2*15*2*14
3rd 2 seats 13 couples times 2 ways of seating a couple = 2*13==>2*15*2*14*2*13
. . . . .
N = 15!*2^15 = 1,307,674,368,000*32,768 = 42,849,873,690,624

(b) This condition rules out all but one factor of 2, so
N = 2*15! = 2,615,348,736,000