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ale_sicheri
A college cafeteria offers three sizes of pizza - small, medium, or large. With a small pizza, up to one topping is included without additional charge; for a medium or large, up to two different toppings are included without additional charge. The cafeteria offers two meat toppings - pepperoni and sausage. If the cafeteria offers N other toppings, then how many ways can someone order a pizza - choosing a size and up to the maximum number of toppings - without having to pay extra? Assume that double toppings are not an option.

A. 3/2N^2 + 9/2N + 3

B. 2N^2 + 7N + 6

C. N^2 + 7/2N + 3

D. 3N^2 + 9N + 6

E. N^2 + 4N + 4

Please let me know your reasoning as I cannot find a proper answer to this question anywhere.
Thanks

There are two possible scenario in which small pizza can be ordered.

1) When 1 topping is selected from meat sausage
2) When 1 topping is selected from other N toppings


2c1+ Nc1 = 2+N

Now, let's look at the way to order small or large pizza

1) when both the toppings are ordered from meat
2) When 1 topping is ordered from meat, and other from remaining N
3) When both the toppings are ordered from N other toppings

Also, since both medium, and large pizza offers two toppings, and the number of ways to order the pizza is identical hence we have to multiple the above mentioned scenarios with 2.
(2c2)*2 + (Nc1*2c1)*2 + (Nc2)*2
=2+4N+N(N-1)
=2+3N+N^2

Thus total number of ways to order pizza = N^2+3N+2+N+2 = N^2+4N+4

IMO E
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Can you let me know why we are excluding case in which no toppings will be selected for all small, medium and large pizza. As per me, it should be included as that will also incur no additional charge.
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hey,
i have gone through the solution above but feel that the answer is wrong. The number of ways of selecting the toppings for both medium and large pizza will be the same but while trying to find the total number of ways, won't we add (n+2)(n+1) twice . Once for large pizza and once for medium s
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