These questions were never answered and I'm stuck! Please Help!

Question 3

Anchor   any adult in seat 1   and  any child in seat 2

We have 3 ways to do the first and  3  ways to do the second

In seat 3 choose any 2 adults

In seat 4 choose any 2 children

In seats 5 and 6 we have one way to choose an adult and one way to choose a child

Total arrangemets  =  3 * 3 * 2 * 2 * 1 * 1 =   36

Question 2

Here's what I get

Choose any 3 of the 4 "suites"  =  C (4,3)

In any one of  of  these choose  any 2 cards of 7  in  one  "suite"  = C(7,2)

And we  can choose any one  of the  three "suites" for  choosing 2  cards of the same  color =  C(3,1)

In the two remaining  suites choose  1 of  these  =  C(2,1)

And in this  suite  choose any 1  of the  7 cards

Total hands =  C(4,3) * C (7,2) * C (3,1) * C( 2,1) * C (7,1)   = 3528

Question 1

Six children are each offered a single scoop of any of 3 flavors of ice cream from the Combinatorial Creamery. In how many ways can each child choose a flavor for their scoop of ice cream so that some flavor of ice cream is selected by exactly three children?

We can select 3 children from any  of the  6  to have the same  flavor  =  C (6,3) =  20  different sets  of  children

And each of these sets  has  three ways to  choose the same  flavor  = C (3, 1) = 3

The remaining 3 children are by default.....one  of then  can choose any  of  the  2  remaining flavors and  this  child can be any one  of the remaining  3...the other 2 children are stuck with the  remaining  flavor not yet selected...so the  total sets here  =  C(3,1) * C(2,1)

So....the total  number of ways = 

C(6,3) * C (3,1) * C(3,1) * C(2,1)  =

20  * 3  *  3 * 2  = 

360  ways

Q3

Obviously, the only way for that to happen is if they sit ACACAC.

We now permutate the children and permutate the adults.

\(3!3!=36\)

But, we need to correct for overcounting. The table has a 3-fold rotational symmetry and a 2-fold one. Therefore:

\(\frac{36}{6}=\boxed{6}\)