Although it may seem like Tosin has many options with 3 ingredients per sandwich and 4 ingredients to choose from, she wants more! She's looking for perfection. The greater variety of sandwiches she offers, the better.
In the table below are a few more ingredient ideas Tosin has for her perfect sandwich. Try to calculate number of combinations possible on your own, before you click on the table to reveal the answer.
| How many unique sandwiches can Tosin make with 4 ingredients per sandwich and 5 ingredients to choose from? | Since she's going to use 4 ingredients in a sandwich, r = 4. She has 5 ingredients to choose from, so n = 5. Therefore, \(\mathsf{ \frac{n!}{r!(n-r)!} = \frac{5!}{4! \times (5-4)!} = \frac{5!}{4! \times 1!} = \frac{5!}{4!} = 5 }\). |
| How many unique sandwiches can Tosin make with 3 ingredients per sandwich and 6 ingredients to choose from? | She's going to use 3 ingredients in a sandwich, so r = 3. Since she can choose from 6 ingredients, n = 6. Thus, \(\mathsf{ \frac{n!}{r!(n-r)!} = \frac{6!}{3!(6-3)!} = \frac{6!}{3! \times 3!} = 20 }\). |
| How many unique sandwiches can Tosin make with 2 ingredients per sandwich and 4 ingredients to choose from? | Here, r = 2 and n = 4. So, \(\mathsf{ \frac{n!}{r!(n-r)!} = \frac{4!}{2! \times 2!} = \frac{24}{4} = 6 }\). |
