[Counting Possibilities] Consider the following pizza deal: 2 pizzas up to 3 top

Question

[Counting Possibilities]

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, c.g., tomatocs 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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