Question

Data Management combinations ?

A burrito restaurant offers 3 different fillings and 4 sizes to choose from. You can choose from 0, or 1, or 2, or 3 fillings for a burrito. If the owner wants to offer more than 2000 possible burritos, how many different sauces must he make available? A customer can choose to have no sauce. You can round up your answer.

Answer 1

`N_("sauce")>=6`

Explanation:

Let's figure out how many choices there are for fillings given that there are `3` different fillings and we can choose `0`, `1`, `2`, or `3` of them at any time.

`N_("fill") = ((3),(0))+((3),(1))+((3),(2))+((3),(3)) = 1+3+3+1 = 8`

From this we see that the number of combinations for any number of choices is the sum of the row of Pascal's triangle. It can be shown that this can be calculated as see this link

`\sum_(0<=k<=n)((n),(k))=2^k`

The number of sizes is `N_("size")=4`, so the number of choices before choosing a sauce is:

`N_("no_sauce") = N_("fill")*N_("size") = 8*4 = 32 = 2^5`

Assuming that we can only choose `0` up to `N_("sauce")` sauces, if we want `N_("burrito")>=2000` then

`N_("burrito") = N_("no_sauce")*2^(N_("sauce"))>=2000`

We also know that there is a number just above `2000` which is a power of `2`, i.e. `2^(11)=2048` so let's re-write in powers of `2`

`2^5*2^(N_("sauce"))>=2^(11)`

Taking `log_2`

`5 + N_("sauce")>=11`

`N_("sauce")>=6`