Choose

 Sometimes mathematics seem complicated, but they really aren't. It can be

a simple matter of mastering definitions. Combinations and permutations are

a case in point.

Below, a combination of three fruit, lemon, apple and lime.

Here, a permutation of these same three pieces of fruit. Permuations are orderings;

these are the possible ways of presenting my three fruit.

*     *     *

Foodie me goes to a shop on a hot day, for sherbet. I want 2 scoops, and order

is not an issue. But there is another possibility here. I can choose two of the same

flavor. It is a combination but with (possibly) repetition.

So, how many choices do I have. If I forego repetition, I have three: Lemon/Apple,

Lemon/Lime, and Apple/Lime. 

n = number of possibilities

r = how many I choose

Combinations and Permutations (mathsisfun.com)

                                                                              

                                                                    *     *     *

The following was suggested to me as a problem:

COPILOT:

Mathematics problem; How can I use these formulas to solve my problem. I go to an ice cream shop that offers 15 flavors. I want to oder three scoops, using two flavours, in a pateern in the bowl od a,b,a. How many choices do I have?

To solve your problem of choosing three scoops of ice cream in a pattern of a, b, a using two flavors from 15 available flavors, you can use the concept of permutations since the order matters.

Here’s the step-by-step process:

Choose flavor ‘a’: You have 15 options.

Choose flavor ‘b’: You have 14 options (since it must be different from ‘a’).

Since the pattern is fixed as a, b, a, you don’t need to calculate further permutations. Simply multiply the choices:

15 (choices for ’a’)×14 (choices for ’b’)=210 combinations

So, you have 210 different ways to choose three scoops of ice cream with two flavors in an a, b, a pattern. Enjoy your ice cream! 🍦

                                                             *     *     *

Both Meta and Gemini present that 210 is it, for choice; and that the order of the scoops is

a settled matter. (This might imply a permutation of 3 for the possible orders of two 'a' and

one 'b'.) It i, in effect, a different problem. I did manage to coax COPILOT with this next question:

                                                             *     *     *

                                                        

How does this change if I am uncertain about the order of the scoops in the bowl. Should I just multiply the 210 by 3?

If you’re uncertain about the order of the scoops, you need to consider all possible arrangements of the flavors. For the pattern a, b, a, there are three possible arrangements: a, b, a, a, a, b, and b, a, a.

However, since the pattern a, b, a is fixed and you want to consider all possible orders, you should calculate the permutations of the two flavors in three positions. The formula for permutations with repetition is:

n1​!⋅n2​!n!​

where ( n ) is the total number of scoops, and ( n_1 ) and ( n_2 ) are the counts of each flavor.

                                                                                        

Now, multiply this by the number of ways to choose the flavors:

210 (ways to choose ’a’ and ’b’)×3 (permutations)=630

So, if you’re uncertain about the order of the scoops, you have 630 different ways to choose and arrange three scoops of ice cream with two flavors. Enjoy your ice cream! 🍦