Hello. In the pizza example, does it take into consideration choosing all 3 of the same toppings? Or is it only in combinations of 01, 0, and 110? Hypothetically, how would we solve that question?

Hey Wilson, Thank you for reaching out! The solution below covers the following situations: 1. All 3 toppings are the same 2. 2 of the toppings are the same and the third one is different 3. All 3 toppings are different You can verify that this is indeed the case through counting. Consider case 1. where all toppings are the same—we have 6 possible outcomes. Next, consider case 2. where two of the toppings are the same and the third one is different, we can have the following outcomes: cheddar cheese, cheddar cheese, X (where X can be wither of the remaining 5 toppings) onions, onions, X green peppers, green peppers, X mushrooms, mushrooms, X pepperoni, pepperoni, X bacon, bacon, X Therefore, case 2. covers 5 x 6 = 30 of all possible outcomes. Finally, case 3. can be solved using the 'combinations without repetition' formula: C(6, 3) = 20. Summing the results from all cases together gives indeed 6 + 30 + 20 = 56 outcomes. Hoe this helps! Kind regards, 365 Hristina

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Wilson Lee

Last answered:

20 Jan 2023

Posted on:

13 Jan 2023

Resolved: Combinations with Repetition

in Probability / Combinations with Repetition

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Hristina Hristova

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Posted on:

20 Jan 2023

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Resolved: Combinations with Repetition

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