3. Consider the following pizza deal 2 pizzas up to 3 toppings on each pizza ·7
ID: 3360057 • Letter: 3
Question
3. Consider the following pizza deal 2 pizzas up to 3 toppings on each pizza ·7 toppings to choose from total cost S10.99 The pizza toppings need not be unique-double or triple toppings of the same kind are allowed; i.e., 2 or 3 of the toppings can be the same. (Note that double or triple toppings are not required, but they are allowed.) A pizza with no toppings also is allowed. The two pizzas may or may not be identical. The arrangement of the toppings on each pizza does not matter; e.g., tomatoes on top of pepperoni is the same as pepperoni on top of tomatoes What is the total number of possibilities for a pizza order in this deal?Explanation / Answer
Please note nCr = n! / [(n-r)!*r!]
Let us take one Pizza. Lets assume the toppings are A, B, C, D, E, F and G.
(a) Number of ways of getting 0 toppings = 1 ways
(b) Number of ways of getting 1 topping = 7C1 = 7 ways ( A, B, C, D, E, F or G)
(c) Number of ways of getting 2 different toppings = 7C2 = 21 ways (AB, AC, AD, AE, AF, AG, BC, BD, BE, BF, BG, CD, CE, CF, CG, DE, DF, DG, EF, EG and FG)
(d) Number of ways of getting 2 similar toppings = 7 (AA, BB, CC, DD, EE, FF and GG)
(e) Number of ways of getting 3 different topping = 7C3 = 35 ways
(f) Number of ways of getting 3 similar toppings = 7 (AAA, BBB, CCC....)
(g) Number of ways of getting 2 similar and 1 different topping = 36
AAB, AAC, AAD, AAE, AAF, AAG = 6 ways with double toppings of A. We will have 6 types for each i.e Double for B will have 6, double for C will have 6...Therefore 6 * 6 = 36
Total Topping possible on 1 pizza = 1 + 7 + 21 + 7 + 35 + 7 + 36 = 114 ways
Therefore there will be 114 different ways for the second pizza as well.
Total Number of ways = 114 * 114 = 12,996 ways
(Why Multiplication: Because we needed to take both pizzas and hence they become dependent events and so we multiply)
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