In order to be photographed, 12 people stand in two rows of 6, one in front of the other. A. In how many ways can this be done if there are no restrictions?
B. Alex and Beth (2 specific people) are in the front row
C. Alex and Beth are in different rows?
What I’ve done so far is that I’ve modelled this as 12 C 6. However, I later tried to multiply it by 2 to signify the two different rows the question poses. I’m not exactly sure how to proceed from here.
Edit: I now have tried the first question in a different way. I know that in order to solve this if it were a single row of 12 people, it would be 12 people in the 1st row, 11 in the second, and so on. However, the question poses 2 different rows of 6 people so I’m once again confused on how to proceed.
Edit 2: Realized that 2 different rows of 6 is essentially the same as 1 row of 12, therefore the answer is still 12!.
Now do not know how to proceed with questions B and C.
Once you’ve picked six people for the back row, you can order the two rows in $6!\cdot 6!$ ways.
A. With no conditions, there are $\binom{12}{6}$ ways to pick the back row, so $6!6!\cdot\binom{12}6=12!$ total ways to arrange people.
B. If Alex and Beth are in the front row, there are $\binom{10}6$ ways to pick the back row with neither Alex nor Beth. So the total is: $6!6!\cdot \binom{10}{6}=6\cdot 5\cdot 10!$
C. If Alex and Beth must be on different rows there are $2\cdot \binom{10}5$ ways to pick the back row, so there are $6!6!\cdot 2\binom{10}5=6^2\cdot 2\cdot 10!$ ways.
A.  first choose the 6 people to stand in the front row.  There are 12 ways to choose the first person, 11 ways to choose the second person, 10 to choose the third, 9 to choose the fourth, 8 to choose the 5th, and 7 to choose the 6th.  That is 12(11)(10)(9)(8)(7)= 12!/6! ways to choose the 6 people to stand in the front row and that automatically determines the 6 who stand in the back row.  There are then 6! ways to order the 6 people in the front row and 6! ways to order the 6 people in the back row.
So there are (12!/6!)(6!)(6!)= (12!)(6!) ways to do this.