https://web2.0calc.com/questions/binomial-help-needed

https://web2.0calc.com/questions/need-counting-help-asap

https://web2.0calc.com/questions/counting_31933

https://web2.0calc.com/questions/3-part-question-please-help-cphill-melody-or-ep

P.S. The already figured out the first part of the 3 part question :)

Keihaku Sep 27, 2023

#1**0 **

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

CPhill Sep 27, 2023

#2**0 **

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

CPhill Sep 27, 2023

#6**0 **

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

CPhill Sep 28, 2023