Ice cream store two topping on ice cream cone. 106 ways choose topping how many toppings(including no toppings)

106P2=106!/104!=106*105=11130

Or, is it 106C2=106!/2!*104!=106*105/2*1=5565?

Which one?

We can imagine the problem like this:

There are "x" toppings to choose from. 

There are \(\binom{x}{2}\) ways to choose the toppings. 

Therefore \(\binom{x}{2}\) = 106 

We can rewrite this as:

\(\frac{x\cdot(x-1)}{2}=106 \)

Then we have to add the choice of "no" toppings. 

\(\frac{x\cdot(x-1)}{2}+1=106 \\ x^2-x-210=0 \\ (x+14)(x−15)=0 \\ x_1=-14,\ x_2=15 \)

Since there cannot be negative toppings, the answer is 15. 

We can double check this.

If there are 15 toppings and you can have no toppings, how many ways can we "top" our ice-cream. 

\(\binom{15}{2}+1=106\), it works!

I hope this helped,

gavin