My Math Forum A fairly basic counting problem that I'm drawing a blank on.

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 January 5th, 2013, 04:17 AM #1 Newbie   Joined: Jan 2013 Posts: 12 Thanks: 0 A fairly basic counting problem that I'm drawing a blank on. Hi all, I hope I'm posting this in the correct forum: The question is this: An ice cream store has 20 different flavours. In how many ways can we order a dozen different ice cream cones, if each cone has 2 different flavours ? If it was just 12 different cones from 20 different flavours, I would go C(20,12) = 125,970 but it's the "if each cone has 2 different flavours that I'm drawing a blank on. I imagine I'll feel like an idiot when it's pointed out, but any help would be much appreciated. Thank you.
 January 5th, 2013, 08:56 AM #2 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 Re: A fairly basic counting problem that I'm drawing a blank i believe you just have to multiply that number with 20*19 because that's the amount of ways you can get 2 different flavours and divide by 2 since all combinations have been taken twice (such as {lemon,vanilla} is the same as {vanilla, lemon} ) so 125,970*190= 23,934,300
 January 5th, 2013, 09:00 AM #3 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: A fairly basic counting problem that I'm drawing a blank $\left(\frac{\binom{20}{1}\binom{19}{1}}{2}\right)^ {12}=\left(\frac{19\cdot 20}{2}\right)^{12}=\left(10\cdot 19\right)^{12}$
 January 5th, 2013, 10:19 AM #4 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: A fairly basic counting problem that I'm drawing a blank This is what I think. First choose 12 from 20, which is 20_C_12. Next, for each cone you have 19 degrees of freedom left so multiply by 19^12. Answer should be (20_C_12) x (19^12).
 January 5th, 2013, 10:55 AM #5 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: A fairly basic counting problem that I'm drawing a blank Sorry, I was in error. Mr Zardoz has it correct
January 5th, 2013, 02:19 PM   #6
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Re: A fairly basic counting problem that I'm drawing a blank

summsies!

The word "different" makes the problem trickier.

Quote:
 An ice cream store has 20 different flavours. In how many ways can we order a dozen [color=blue]different[/color] ice cream cones, if each cone has 2 different flavours?

$\text{Each cone would have a different }pair\text{ of flavors.}$

$\text{With 20 flavors, there are: }\,{20\choose2} \,=\,\frac{20!}{2!\,18!} \,=\,190\text{ different pairs of flavors.}$

$\text{Now select 12 of them: }\:{190\choose12} \:=\:\frac{190!}{12!\,178!}$

January 5th, 2013, 09:23 PM   #7
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Re: A fairly basic counting problem that I'm drawing a blank

Quote:
 Originally Posted by soroban summsies! $\text{Now select 12 of them: }\:{190\choose12} \:=\:\frac{190!}{12!\,178!}$
i calculated that if u need it in numbers.
it should be 3345763705175979360

 January 6th, 2013, 01:12 AM #8 Newbie   Joined: Jan 2013 Posts: 12 Thanks: 0 Re: A fairly basic counting problem that I'm drawing a blank Thanks to all who replied: Soroban, yes it was definitely the wording of different cones each with different flavours that tripped me up. These sneaky Math profs, hey So just to recap so I understand: From 20 different flavours, we're choosing 2 flavous per cone, giving us 190 possible combo's there, and then from that 190 possible combinations, we're making 12 different cones. I think that makes sense. Thanks again to all for your input !
January 6th, 2013, 01:21 AM   #9
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Re: A fairly basic counting problem that I'm drawing a blank

Quote:
 Originally Posted by ZardoZ $\left(\frac{\binom{20}{1}\binom{19}{1}}{2}\right)^ {12}=\left(\frac{19\cdot 20}{2}\right)^{12}=\left(10\cdot 19\right)^{12}$
Zardoz - In your answer, is your line of reasoning: 20 ways to choose the first flavour, 19 ways to choose the second, divide by 2 different flavours = 190 ^12 because there are 12 different cones being made? Both yours and Soroban's solutions make sense to me!

 January 6th, 2013, 02:11 AM #10 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: A fairly basic counting problem that I'm drawing a blank [color=#000000]Soroban's solution is correct![/color]

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