algebra formula

bingobrad

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I asked this question of my ninth grade nephew yesterday at dinner. I can't get the formula figured out and I hope you can help.
If there are 3 choices (i.e chocolate, vanilla, and strawberry ice cream) and you can have any combination of the 3 individually, together, or you can have all 3, how many possible combinations are there? There would be 7 possible combinations according to my analysis. If there are two choices there would be 3 possible combinations, if there are four choices there are 15 possible combinations, and if there are five choices there are 30 possible combinations according to my analysis. What would be the algebraic formula to solve this problem for any specific number of choices?
 
bingobrad said:
I asked this question of my ninth grade nephew yesterday at dinner. I can't get the formula figured out and I hope you can help.
If there are 3 choices (i.e chocolate, vanilla, and strawberry ice cream) and you can have any combination of the 3 individually, together, or you can have all 3, how many possible combinations are there? There would be 7 possible combinations according to my analysis. If there are two choices there would be 3 possible combinations, if there are four choices there are 15 possible combinations, and if there are five choices there are 30 possible combinations according to my analysis. What would be the algebraic formula to solve this problem for any specific number of choices?

Yes - if we exclude no-topping as a choice - then the answer is (2[sup:tnbrdkt2]n[/sup:tnbrdkt2] - 1)

The actual formula is:

\(\displaystyle \sum_{k=0}^{n}\frac{n!}{k!\left(n-k\right)!}=2^{n}\)

or

\(\displaystyle \sum_{k=1}^{n}\frac{n!}{k!\left(n-k\right)!}=2^{n} \ \ - \ \ 1\)


Which by the way tells us that the #of choices with 5 toppings should be 31.
 
Thank you for that Khan. You made it seem so easy. I tried every formula combination I could think of without any luck.
 
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