1. ## probability

How many different ways can a family of 6 sit in a row at the movies if the 2 youngest children must sit next to each other?

2. First, there are 5 ways the 2 children can be sit. In each case the rest of the family can be sit in 4! different ways. So I think the answer is 5.4!. And it's equal to 120. (Which is 5! So now I see that we could have think the problem as if the 2 children were one person of the family and not 2).

3. Hello, Help09!

I got a different answer . . .

How many different ways can a family of 6 sit in a row at the movies
if the 2 youngest children must sit next to each other?
Duct-tape the two youngest together.
. . Then we have five "people" to arrange: . $\boxed{AB}\;C\;D\;E\;F$
And they can be arranged in: . $5! = 120$ ways.

But in each of those arrangements, $\boxed{AB}$ can be replaced by $\boxed{BA}$

Therefore, there are: . $2 \times 120 \:=\:{\color{blue}240}$ ways.

4. Let these be the seats

--- --- --- --- --- ---

We first fix the first two seats for the youngest children

[--- ---] --- --- --- ---

There are 4! ways of filling the other 4 seats

Similarly for the cases,

--- [--- ---] --- --- ---

... so on for 5 cases, each of which contributes a 4!. So the total number of ways is 5*4!=120 * 2 = 240 (where 2 children can be swapped)

In general for n seats and k people in connesecutive seats,
[(n-k+1)(n-k)!]k! = (n-k+1)!k!