#1
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A dumb bar bet...
Hey all
I was hoping for some assistance from the statisticians to settle a dumb bar bet. I run a bar that carries 93 different liquors and cordials and 12 non-alcoholic mixers. We were debating how many drinks we could actually create (some of them being completely undrinkable not withstanding) assuming no drink would have more than 5 alcoholic and 3 non-alcoholic ingredients. Any way I can do this math? My Thanks AAeyes |
#2
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Re: A dumb bar bet...
93C5 * 12C3, no?
I don't have a calculator on me so I don't know the answer or if that's the actual equation to use. I'm likely way off. |
#3
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Re: A dumb bar bet...
correct equation or not, that's the best avatar in the world...
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#4
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Re: A dumb bar bet...
Better than Jakethebake's:
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#5
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Re: A dumb bar bet...
well before jumping into combinations lets think about it.
each alcoholic beverage can be mixed w/ either 1,2,or upt eo 3 mixers. so we have 93*3 immediately for the 1 alcoholic drink + all possible mixers. now, for 2 alcoholic drinks we do 93C2 for the total # of 2 combinations out of 93 objects. so there's 4278 different ways to mix 2 alcoholic drinks. again we take this # and *3 to get the total # of 2 alcoholic drink mixes. now for 3,4,5 its the same thing, ... Total= 93C1*3 + 93C2*3 + 93C3*3 + 93C4*3 + 93C5*3= 165,075,465 so there you have it over 165 million different drinks. unless i made a mistake which i may have. Barron |
#6
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Re: A dumb bar bet...
My answer is 16,123,871,372,127 drinks
94*94*94*94*94 7,339,040,224 x 13*13*13 2,197 = 16,123,871,372,128 - 1 (because a drink with no alcohol and no mixer isnt a drink) Assumptions: 7-up with no alchohol is OK Scotch with no mixer is OK Double scotch with no mixer is not the above. |
#7
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Re: A dumb bar bet...
That's what I was thinking at first, but I have to figure on the different combinations of mixers too and I don't know the equation. I think it's more than 165 million because I have to figure on each combo of liquor with every combo of mixer. So vodka cranberry + vodka oj + vodka cranberry oj + vodka soda + vodka cran soda + vodka cran OJ soda...... and I can't figure in all the various combinations. I think it should come out to billions of possibilities....
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#8
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Re: A dumb bar bet...
Sorry, didn't see you weren't yet jumping into different combinations...
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#9
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Re: A dumb bar bet...
So the formula is that simple? The number of liquors to the power of however many times I can use them * etc. etc.? somehow I thought it would be more difficult, but then again I'm not much of a statistician.
Many Thanks, AAeyes |
#10
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Re: A dumb bar bet...
That's the answer for drinks with exactly 5 liquors and 3 mixers. The OP asked for all drinks, which includes those with just 3 liquors and 2 mixers, etc. So the real answer is:
93C5*12C3 + 93C4*12C3 + 93C3*12C3 + 93C2*12C3 + 93C1*12C3 + 93C5*12C2 + 93C4*12C2 + 93C3*12C2 + 93C2*12C2 + 93C1*12C2 + 93C5*12C1 + 93C4*12C1 + 93C3*12C1 + 93C2*12C1 + 93C1*12C1 + 93C5*12C0 + 93C4*12C0 + 93C3*12C0 + 93C2*12C0 = 16452521345 Edit: I'm not counting any drink without alcohol as a drink. |
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