Create Your Sundae - 45 Combination Options for Two Scoops

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The solution is similar to the ice cream problem, where the first person has 9 possible handshakes, the second person has 8 possible handshakes (since they already shook hands with the first person), and so on. So the total number of handshakes is 9+8+7+6+5+4+3+2+1=45. In general, for n people, the number of handshakes is n(n-1)/2.
  • #1
chemistry1
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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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  • #2
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?
 
  • #3
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 !
 
  • #4
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 [itex]^{10}C_2[/itex].
 
  • #5
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.
 
  • #6
Ok, Thank you again. I didn't know the standard way...^^
 
  • #7
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.
 
  • #8
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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