A dumb bar bet...

[AAeyes]

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

[tbach24]

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.

[AAeyes]

correct equation or not, that's the best avatar in the world...

[tbach24]

Better than Jakethebake's:

[DcifrThs]

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

[NutzyClutz]

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.

[AAeyes]

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....

[AAeyes]

Sorry, didn't see you weren't yet jumping into different combinations...

[AAeyes]

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

[DougOzzzz]

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.