9 people are to sit in 3 rows of 3 people.
'Jim' and 'Bob' must be in the same row.
How many ways can this be arranged?
I'm not sure how to go about solving this.
I see it as one row has to have 2 people together, but the row needs 3 total, so I would guess:
3!
You then have to place this row in one of the 3 rows, so it becomes 3! x 3
You then have to arrange the next 6 people in the remaining seats, so 6!.
So my guess is:
3! x 3 x 6! = 12,960 ways
However, I feel that this is wrong.
Best Answer
You forgot to account for the third person who sits in the same row as Jim and Bob.
Method 1: Here is a modified version of your solution.
There are $3$ ways to choose the row in which Jim and Bob sit. There are $7$ ways to choose the third person who sits in that row. There are $3!$ ways to arrange the people in that row. There are $6!$ ways to arrange the remaining six people in the other six seats. $$3 \cdot 7 \cdot 3! \cdot 6!$$
Method 2: We seat Jim and Bob, then sit the remaining people.
There are nine ways to seat Jim. Since Bob must sit in the same row, there are two ways to sit Bob in one of the remaining seats in that row. The seven remaining people can be seated in the seven remaining seats in $7!$ ways. Hence, there are $$9 \cdot 2 \cdot 7!$$ permissible seating arrangements.