# Summer-Math-Quiz: Cocktails!!

Since I am just coming back from a one month trip, where I visited a couple of wonderful places in north + south America, I am feeling inspired to first finally give the solutions to the Summer-Quiz posts from last summer, when I was lucky to travel from Bavaria through Tirol to Italy, and continued by the Italian-French Riviera to Cap Martin, Juan Les Pins, Nice and Cannes, coming back through the Lago Maggiore, St. Moritz and Lichtenstein, to Lindau in Bavaria.

I used those weeks to get offline math inspirations and thought about a couple of nice and easy questions anyone could make during a sunny summer… Like in…

Garda Lake, Italy: Oh! Summer nights!!… The higher temperatures make you sooooo thirsty 😉 and there are sooooo many different drinks at the menus of the bars!… Actually, this is not just a completely hypothetical problem ;). On the other hand, I found it can wonderfully show different approaches of how we think and work in math. Thus, I post this to show:

1. The most important & difficult thing in math usually is: UNDERSTANDING THE QUESTION.
2. Different approaches to solve a problem.
3. I found a so pretty formula 🙂 !
• Quiz: Say you want to try 2 drinks each of the 3 nights you are planing to go for drinks… In how many ways can you do that, if you choose the cocktails among those you think are your 7 favorite ones?

Note, since this is a “real life question”, it is very vague! Then, let us find out answers for possible accurate questions. What should be clear is: You have to pick 6 among 7 drinks. Nice! … 😉 … But you miss 1 😦 … You don’t repeat drinks, since you want to try as many as possible!… There are soooo many…

Q1: Do you want to know ALL the different ways you can choose 6 drinks among 7? That means you care about the way you choose them & the order you drink them!… Or…

Q2: You want to know ONLY how many different choices of 6 among 7 drinks are there, without caring about the order?… Or…

Q3: Do you want to know the different ways you can pick the 3 “pairs” of drinks among 7? (I never thought about this before and the answer is very nice 🙂 )

Answer Q1: You pick the 1st drink among the 7 you like, then the 2nd among the 6 you still like, then the third and so on. For the 1st you have 7 alternatives, for the 2nd you still have 6 choices, for the 3rd just 5, for the next 4, for the next 3, for the last drink only 2 alternatives. This is a total of 2 x 3 x 4 x 5 x 6 x 7 (= 7!) = 5040 !!! (Quite many different ways to choose & order 6 out of 7 isn’t it 😉 )… (See Combinatorics – Wolfram MathWorld or Probability and Combinatorics – KHANACADEMY).

Answer Q2: If you ONLY want to know how many different “sets” of 6 drinks can you choose (in a set the order is irrelevant!). Then the question is only in how many ways you pick 6 things among 7, which actually is the same as to (don’t 😉 ) pick 1 among 7. This is just 7 ways! (Common sense!!!) (And you can check it with the formulas at the end: 7! / 6! (7 – 6)! = 7! / 6! = 7! / 1! (7 – 1)! = 7).

Answer Q3: In how many ways you can pick 3 PAIRS of drinks among 7 you like?

Well, first you have to choose 6 among the 7. You do that in 7 different ways. Then for each way you have to count the number of different configurations of pairs. The total number is 7 multiplied by this number!

Thus, you try to count in how many ways you can make 3 pairs of drinks with 6 already chosen. The nice thing I found is, that without knowing any formulas, you can find a recurrence (which you can prove by induction! 🙂 ). You may think this way:

How many PAIRS can you make with 2 drinks:  #_2 = 1

How many PAIRS can you make with 4 drinks:  #_4 = ?… Call the drinks just 1, 2, 3 and 4 and the pairs 1,2 ; 1,3 ; 1,4;  2,3 ; 2,4 and 3,4. (There are 6 different pairs! Note i,j = j,i ). Now count!:

Night 1: You could have the pair 1,2 or 1,3 or 1,4  (3 possibilities). Or the pair 2,3 or 2,4 (2 possibilities). Or the pair 3,4 (1 possibility). A total of 3 + 2 + 1 = 6 = 4 x 3 / 2  (Since 1 + 2 + 3 + … + N = N (N + 1) / 2  … By the genius of Gauß!)

Night 2: You have only one pair left (1 Possibility)

Total of pair configurations:  #_4 = 6 x 1 = 6   ( = 4 x 3 / 2 😉 )

How many PAIRS can you make with 6 drinks:  #_6 = ?…

Night 1: You can choose the pair of drinks 1,2 or 1,3 or … or 1,6 (5 possibilities). Or the pair 2,3 or 2,4 or … or 2,6 (4 possibilities). Or the pair 3,4 or … or 3,6 (3 possibilities). Or … … … Or the pair 5,6 (1possibility). This is a total of  5 + 4 + 3 + 2 + 1 = 15 = 6 x 5 / 2.

Night 2: You have only 4 cocktails, and you already know that there are  #_4 = 6 = 3 x 4 / 2 different configurations of 2 pairs. So now the total number with 6 pairs is

#_6 = (6 x 5 / 2) x #_4 = (6 x 5 / 2) x (4 x 3  / 2) = 360 / 4 = 90.

And the final answer is: 7 x 90 = 630 … (Wow! But much less than 5040. Actually this is 1/8 of 5040).

You can check your result using what follows. Let’s recall the definition of the notation N! (called “N Factorial”):

0! := 1 and for any natural number N, N! := 1 x 2 x 3 x … x N.

And the number of ways you can pick N things (“sets” of N things) among M different things is:

M! / N! (M – N)!

Thus, what you do is, you choose 2 from 7, then choose again 2 from the 5 left and finally you choose 2 from the 3 left. This gives

7! / 2! (7 – 2)! x  5! / 2 ! (5 – 2)! x 3 ! / 2 ! (3 – 2)! =

7! / 2! x 5! x  5! / 2 ! 3! x 3 ! / 2 ! 1! = 5040 / 8 = 630. (Yes!)

Remark: Note that with the recurrence method used to count the different ways to pick pairs, you can very easily show by induction that since

#_2 = 1 = 2 x 1 / 2 = 2! / 2^1

#_4 = 4 x 3 / 2 = 4 x 3 x 2 / 4 = 4! / 2^2

and

#_6 = (6 x 5 / 2) x (4 x 3  / 2) = 6 x 5 x 4 x 3 x 2 / 8 = 6! / 2^3

Then. the number #_2N of ways of choosing N pairs among 2N things is

#_2N = (2N)! / 2^N

… 😉 … We ❤ Math…

Moreover, we could have though about this formula immediately, since what we really do is considering ALL possible orders of 2N things and forgetting about ALL the possible orders within the N pairs : This is (2N)! divided by 2^N. (See also this page and this )

Just 4 Fun…