how many different family night possibilities are there? (info in description - pls help!!!)

Question

Can someone please help me with this question? I don't understand how to do it and my teacher explained it but it still makes no sense to me. If it helps, we have been using permutations etc on the calculator to figure out more simple problems.

The Jacksons are planning their next family night. They always have dinner out somewhere and then do something fun together. There are 2 adults and 6 children in the family. Each family member is allowed 3 meal suggestions, and each child is allowed 2 activity suggestions. Assuming no family members choose the same thing, how many different family night possibilities are there?

Answer 1

8 people x 3 meals x 6 children x 2 activities = 288 possibilities.

You might be tempted to add them 8x3 + 6x2 = 24+12=36 But try a more simple problem, such as 1 adult & 1 child choosing among 3 meals and 1 child choosing among 2 activities. You'd get 2x3 meals and 2 activities. That's 12 possibilities, 2x3x2 = 12, not 2x3 + 1x2 = 8.

Here's the possibilities with just the adult choosing 3 meals

Meal 1 with Activity 1

Meal 1 with Activity 2

Meal 2 with Activity 1

Meal 2 with Activity 2

Meal 3 with Activity f

Meal 3 with Activity 2

That's 6 possibilities just for the adult choices of 3 meals, and another 6 for the 3 different meals chosen by the child. That's a total 12 possibilities.

Unless you count the order, whether meal is last or first, then double it to 576 but the meal seems to go first in the problem.

But still that assumes the entire group eat and do everything together

If you split them up in groups of 0 to 8, and count as additional possibilities every different child in the group. You probably don't want children in a group without an adult supervising them, so that may mean 2 subgroups, each with one adult and anywhere from 0 to 6 children in each subgroup.

You can do the combination formula nCr = nPr/r² for r from n to 0. Add up all those different combinations such as 1 Adult going alone the 2nd adult with all 6 kids, 1 adult with 1 kid, the other with 5, etc.

and varying which children are in the 5 or 4 group, you get a far more possibilities.

If you calculate 6C6, 6C5....6C0 that gives all combinations of the children for subgroups splitting up the 6 children. Add those numbers: 1+6+15+20+15+6+1=64. That's the 7th row of Pascal's triangle, or the coefficients of the terms of expanding (a+b)⁶

Multiply 64 x 576 = 36864 different family night activities, or just 64x288 = 18432, if you do the meal first only. However each of these 64 different ways to split up for the evening will duplicate other subgroups' meals & activities.

the question is a little ambiguous, and it's not clear what answer is intended,

but this may be the only way to use permutations or combinations to get an answer

IF you don't have a problem that somehow uses permutations or combinations, the best answer looks like 8x3x6x2 = 288

Answer 2

I don't think permutations work here.

There are 2 adults and 6 children that makes 8. Each is allowed 3 meal suggestions so 8x3 = 24.

Each child is allowed 2 each activity suggestions which is 12.

So 24 + 12 = 36 possibilities.

If they would have stated that there are x number of places to eat and y activities, how many combination could they choose from, or something like that, combination/permutations might apply.