Tosin decides that she should only use one of each ingredient. However, she does wonder about how many unique meals she can create. At her deli, every sandwich comes with a choice of 3 sides. The customer can choose from chips, cookies, pickles, or fruit salad. Customers take however many of each side they want, as long as they end up with a total of 3.
Using the equation for combinations with repetition, how many combinations of sides exist?
In this case, n = 4 and r = 3. Therefore:
\(\mathsf{ \frac{(r+n-1)!}{r!(n-1)!} = \frac{(3+4-1)!}{(3!)(3!)} = \frac{6!}{(3! \times 3!)} = 20 }\)
Tosin wonders how increasing or decreasing the number of sides affects the potential number of combinations. See if you can solve for the number of combinations given the scenarios below. Try to work each problem in your notebook before you click on the table to reveal the answer. If you get stuck or get the incorrect answer, go back to the previous page and watch the video again.
| If Tosin increases the number of sides each customer can choose to 4, how many combinations exist? | Now n = 4 and r = 4. So, \(\mathsf{ \frac{(4+4-1)!}{(4!) \times (4-1)!} = \frac{7!}{4! × 3!} = 35 }\). |
| If Tosin allows the customers to have only two sides, how many combination are there? | Here, n = 4 and r = 2. So, \(\mathsf{ \frac{5!}{2! \times 3!} = 10 }\). |
| If Tosin adds coleslaw as a potential side choice and leaves the choice of sides at 3, how many combinations are there? | n = 5 and r = 3. Using the equation, we end up with 35 possible combinations. |
