You can put this solution on YOUR website!
The first is a combination problem.
As a rule you can state:
One can combine the items from a list "A" with the items from the list "B"
in a x b ways if there are a items in list A and b items in list b.
This promblem gives us 3 lists.
A good way to look at it, I think, is to follow the rule for 2 lists,
then call the result a list and apply the rule again with the third
list. That way you don't need another rule.
So, let's do it.
Soup can be combined with 4 different choices from the 2nd list
Salad can be combined with 4 different choices from the 2nd list
Cottage Cheese can be combined with 4 different choices from the 2nd list
Clole Slaw can be combined with 4 different choices from the 2nd list
4 + 4 + 4 + 4 = 4 x 4 = 16
That proves the rule and gives us our next list which looks like:
soup .. beef
soup .. pork
soup .. chicken
soup .. pasta
salad .. beef
salad .. pork
salad .. chicken
salad .. pasta
cottage cheese .. beef
We've got 16 items in this new list and we want to combine these 16 items
with 3 items of the problem list, sherbet, pudding, ice cream.
Apply the rule:
16 x 3 = 48 and that is the answer
for the 2nd part, I draw what are called Venn diagrams. they look like
circles that overlap partially to show things that the populations
of thing in the circles have in common.
Just with logic, if I have a circle inside of which are students that like baseball, and I have another circle inside of which are students that like
basketball, then students that don't like either have got to be outside
23 from the class like basketball, 15 like baseball, 12 like both.
If 12 like both, and 15 like baseball, then 3 must ONLY like baseball.
If I combine the 23 who like basketball (knowing there are 12 of them
who like both, but not caring about that) with 3 who only like
baseball, then I get 26 students who lie either or both.
That leaves 2 who don't like either.