Question

Suppose some kid has abundant supply of soda of 3 flavors (flavor A, B, and C). How many possible ways can he mix the soda?  

I was thinking 3^3 but then that cant be right as he can mix them just 2 ways as well. So then I was thinking 3^3 + 3(2^2) but 1) im not sure that would get me the correct answer (though im pretty sure it would) and 2) I know there is a better way to do this problem. Especially with a TI-84 with the MATH->PROB functions.

Anyone got some simple answers? I'm kind of a novice at probability and statistics at this point. Any help would be greatly appreciated. Thanks.

Answer 1

A mix means at least two flavors must be together.

Mixing can be of 2 types as follows:

1) Mix all 3 flavors

2) Mix any 2 flavors

Mixing 3 flavors using 3 flavors can be done in C(3, 3) =1 way. ​​​​​​

Mixing any 2 flavors using 3 flavors can be done in C(3, 2) =3 ways.

Thus, the number of possible ways he can mix the soda =1 way + 3 ways =4 ways.

4 ways of mixing soda are:

ABC,

AB,

AC,

BC.

(mixing B and C is same as mixing C and B. So, it is taken only once. Order is not important here. Similarly, others).

Note: C(n, r) =Number of combinations of selecting "r" number of items out of "n" number of items. C(n, r) =n!/[(n-r)!*r!]

"If order is important, we use permutation formula. If order is not important, we use combination formula".