At her favorite ice cream store, Saga can put up to two toppings on an ice cream cone. If there are 106 different ways she can choose toppings (including the choice of no toppings), how many different toppings are there?

rohitaop
May 18, 2018

#1**+1 **

We have

The choice of any two toppings plus 1 with no toppings = 106

C(n, 2) + 1 = 106

C(n ,2) = 105

We need to solve this

n!

________ = 105

(n - 2)! 2!

n!

______ = 105 * 2!

( n - 2)!

(n) (n - 1) = 210

n^2 - n - 210 = 0 factor

(n -15) ( n + 14) = 0

Setting both factors to 0 and solving for n produces n = 15 or n = -14

Since n is positive....there are 15 toppings

CPhill
May 18, 2018

#2**0 **

However, you didn't account for the "up to 2 toppings" Therefore, there can also be one: 15 choose 1= 15 so you subtract:

105-15=90. 14 choose 2 is 90, after adding 1 you get 91.

So the final answer should be 14

Guest May 20, 2018

#3**0 **

Well, what do you know? It looks like Mr. BB got one right, and even caught a mistake by CPhil. Gingerale will be surprised. Maybe she will eat more crow. LMAO!!

Guest May 20, 2018

#4**0 **

Mr. BB may have caught a mistake by Sir CPhill, but he didn’t get the solution quite right (no surprises, here). For one, 14 choose 2 is NOT 90, it’s 91. He didn’t set up an equation for the solution; he solved this intuitively and added some blarney slop, which is his usual method.

The equation is different from the one presented by Sir CPhill. It will look like this:

\(\dfrac{n !}{2!( n-2)!} + \dfrac{n !}{1!(n-1)!} + 1 = 106 \leftarrow \small \text{ solve for n }\\\)

This is much more difficult to solve; it’s an educated guessing process. For small sums, like this one, it’s easy; for large sums, a perusal of Pascal’s triangle would speed up the search, or a computer algorithm.

Anyway, I eat crow frequently; we chimps like them.

^{A certain, annoying troll should read his messages more often.}

GA

GingerAle
May 20, 2018