Create Your Sundae - 45 Combination Options for Two Scoops

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An ice cream shop offers ten flavors, and the discussion revolves around calculating the number of ways to choose two different flavors for a sundae. The initial approach suggested using a summation method, but it was clarified that the standard calculation involves combinations, specifically 10 choose 2, resulting in 45 unique combinations. The reasoning highlights that while there are 90 arrangements considering order, only half of those are valid since the order of scoops does not matter. The conversation also draws parallels to other combinatorial problems, such as counting handshakes among a group. Overall, the discussion emphasizes understanding combinations over permutations for accurate calculations.
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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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