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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?

QuestionBoi  May 12, 2018
 #1
avatar+2968 
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106P2=106!/104!=106*105=11130

 

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

 

Which one?

smileysmiley

tertre  May 12, 2018
 #2
avatar+930 
+2

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

GYanggg  May 12, 2018

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