P.S. The already figured out the first part of the 3 part question :)
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
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
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 =
Obviously, the only way for that to happen is if they sit ACACAC.
We now permutate the children and permutate the adults.
But, we need to correct for overcounting. The table has a 3-fold rotational symmetry and a 2-fold one. Therefore: