A mocktail has to be prepared from 3 kids of whiskies, 4

mbahope, could u please explain why are you adding? I couldnt understand the steps.

Mbahope, you're right. That's the smartest way to combine all the possible sums. I was trying to assess sigma(i) iCj w/o any success...

3255

Here is some explanation -

The key here is "atleast one from each". So,

For whiskies: number of ways of selecting 1 + number of ways of selecting 2 + number of ways of selecting 3 = 3C1 +3C2+3C3 = 7

For sodas: number of ways of selecting 1 + number of ways of selecting 2 + number of ways of selecting 3 + number of ways of selecting 4 = 4C1+4C2+4C3+4C4 = 15

For fruit juices: using same logic = 31

Now multiply all 3 to get the answer.


ANOTHER APPROACH:

FYI: "at least x" = All possibilities - [equal to (x-1)]
So whiskies = all selections - ways of selecting 0 number = 2^3-1 = 7
Sodas = 2^4 - 1 = 16-1=15
Fruit juice = 2^5 - 1 = 31

I hope it is clear.

Thanks aspire got it now!

Aspire2005. Thank you very much for your explanation! I had no clue how to do this one Fantastic approach and thanks for the detailed explanation.

A very interesting approach! I am not sure how we got to 2^3 (2^4...) and subtracting 1 from each without first doing it the mbahope way above?

Thanks aspire.
I have hard time explaining some things i do in words.
I guess thats something i really need to work on.
I've started trying to do that now.
But i'm glad i dont have to do it for this question now

thanks aspire. thats a very good picture of how to go about such problems..