The student council is ordering pizza for their next meeting.
There are 20 council members, 7 of whom are vegetarian. A 
committee of 3 will order 6 pizzas from a pizza shop that has 
a special price for large pizzas with up to three toppings. 
The shop offers 10 different toppings.

a) How many different pizza committes can the council choose 
if there must be at least 1 vegetarian and 1 non vegetarian 
on the committe?

There are 7 vegetarians and 13 non-vegetarians.

The committee either must consist of 1 vegetarian and 2 non-vegetarians or
2 vegetarians and 1 non vegetarian.

The number of committees consisting of 1 vegetarian and 2 non-vegetarians
is (7C1)(13C2)=7*78=546

The number of committees consisting of 2 vegetarians and 1 non-vegetarian
is  (7C2)(13C1)=21*13=273

Therefore total number of acceptable committes is 546 + 273 = 819 

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b) in how many ways could the committee choose up to 3 toppings for a pizza?

That says for 'a' pizza, that is, just ONE of the pizzas, not all 6 of them.

If they choose exactly 1 topping, that "10 choose 1" or 

If they choose exactly 2 toppings, that "10 choose 2" or 

If they choose exactly 3 toppings, that "10 choose 3" or 

That gives a total of  ways to order just one pizza.

-----------------------

c) The committee decide to order each topping exactly once and to have
at least 1 topping on each pizza. describe the different cases possible?

It doesn't say calculate, but just "describe" them.  So we can have three
cases:

Case 1:
2 pizzas with exactly 1 topping each and 4 pizzas with exactly 2 toppings
each. 

Case 2:
3 pizzas with exactly 1 topping each, 2 pizzas with exactly 2 toppings each,
and 1 pizza with exactly 3 toppings each.

Case 3:
4 pizza with exactly 1 topping each and 2 pizzas with exactly 3 toppings
each. 

--

d) for one of the cases determine the number of ways of choosing and
distributing the 10 toppings.

It just says for ONE of the cases, so I suppose we can pick whichever
case we wish.

Case 1:
2 pizzas with exactly 1 topping each and 4 pizzas with exactly 2 toppings
each.

That's difficult to avoid counting the same things more than once!
But we can do it, but I won't explain it entirely:



The division by 4! was necessary to avoid counting the same things more
than once. 

Case 2:
3 pizzas with exactly 1 topping each, 2 pizzas with exactly 2 toppings 
each, and 1 pizza with exactly 3 toppings each.

That's also difficult to avoid counting the same things more than once!
But we can do it, but I won't explain it entirely:



The division by 2! was necessary to avoid counting the same things more
than once.

Case 3:
4 pizza with exactly 1 topping each and 2 pizzas with exactly 3 toppings 
each.

That's not too difficult to avoid counting the same thing twice.  So I'll
pick that one to explain entirely:  

First we choose a pizza with 3 toppings.  That's 10C3.  That leaves 7
toppings for the other pizza with three toppings. That's 7C3. That would
be (10C3)(7C3)=120*35=4200.  However this counts ever case twice because:

Take the case where we chose pepperoni, mushrooms and olives for the
first pizza, and onions, anchovies, and sausage for the second pizza.
That would be the same as if we had chosen onions, anchovies, and sausage
for the first pizza and pepperoni, mushrooms and olives for the second 
pizza.

Therefore we must divide the 4200 by 2 to get the correct count.

4200÷2 = 2100.

So there are 2100 ways to get the number of ways we could have case 3.

We don't have to consider the 4 pizzas with 1 topping each because these
will just be 4C4 or 1 way for the remaining 4 toppings to go on the
remaining 4 pizzas. 

Edwin