Create Your Sundae - 45 Combination Options for Two Scoops

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SUMMARY

The discussion centers on calculating the number of unique two-scoop sundaes from ten different ice cream flavors, ensuring that both scoops are of different flavors. The correct calculation method involves using the combination formula, specifically 10 choose 2, denoted as ^{10}C_2, which results in 45 unique combinations. The reasoning provided highlights the importance of considering the order of scoops and the necessity to divide by two to account for the indistinguishability of the scoops.

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Hi,
An ice cream shop sells ten different flavors of ice cream. You order a two-
scoop sundae. In how many ways can you choose the flavors for the sundae if
the two scoops in the sundae are different flavors?

That's what I did :

9+(9-1)+(9-2)+...+(9-9)=45

Would this be any good ?

Thank you!
 
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chemistry1 said:
Hi,
An ice cream shop sells ten different flavors of ice cream. You order a two-
scoop sundae. In how many ways can you choose the flavors for the sundae if
the two scoops in the sundae are different flavors?

That's what I did :

9+(9-1)+(9-2)+...+(9-9)=45

Would this be any good ?

Thank you!

It might be if you explain your reasoning. It's not the usual way to write it. Why do you think it's correct?
 
Dick said:
It might be if you explain your reasoning. It's not the usual way to write it. Why do you think it's correct?

Well, we need two different sorts, so we have a total combination of 9 for 10 different flavour.

For example, if we consider each flavour a number, then : (1,2) (1,3) ... til (1,10)

Now, if we want to see for the second flavour, we already did the combination (1,2), so we must substract 1 combination from 9.

so we have 9+(9-1) and we continue this way.

I hope that's what you're seeking !
 
Your reasoning is solid and your answer is correct. It may also be helpful for you to know that the answer is 10 choose 2, which is denoted by ^{10}C_2.
 
chemistry1 said:
Well, we need two different sorts, so we have a total combination of 9 for 10 different flavour.

For example, if we consider each flavour a number, then : (1,2) (1,3) ... til (1,10)

Now, if we want to see for the second flavour, we already did the combination (1,2), so we must substract 1 combination from 9.

so we have 9+(9-1) and we continue this way.

I hope that's what you're seeking !

Yes, that's what I'm seeking and yes, it's correct. There is a more standard way to do this. See http://en.wikipedia.org/wiki/Combination That would let you write it as 10*9/2=45. Same number.
 
Ok, Thank you again. I didn't know the standard way...^^
 
The "C" notation seems to be out of fashion, at least for starting out.
John Allen Paulos' influence I suspect.

The usual approach seems to be to argue there are 10 possibilities for the 1st scoop, and, since the second scoop has to be different, there are 9 remaining ... for 10x9=90 possible sundaes with 2 different flavors... however, that counts say "vanilla+ chocolate" as different from "chocolate+vanilla" ... since the order does not matter, then the required value is half that: 90/2=45.
http://betterexplained.com/articles/easy-permutations-and-combinations/

That was quite interesting reasoning though, I'd like to see that applied to combinations of 3 scoops.
 
This is similar to the problem of counting handshakes: if 10 people shake hands such that every person shakes hands with every other person exactly once, how many handshakes take place?
 

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