Last answered:

20 Jan 2023

Posted on:

13 Jan 2023


Resolved: Combinations with Repetition

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?

1 answers ( 1 marked as helpful)
Posted on:

20 Jan 2023


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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