math – update

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Day #14: Cocktails = Combinatorics!

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l-g-c2
Rosé Champagner

Q: Suppose you will spend a 3-days-sommer-weekend at some warm & wonderful place – like Garda Lake (Italy), or Venice  Beach (Miami), or Pucon (Chile), … – . And say, you want to drink – only – 2 cocktails each and every of the 3 days you´ll be there! Now, if you choose your cocktails among those you think are your 7 favorite ones, and you don´t repeat drinks, can you tell in how many – different – ways you could enjoy them??… Hint: There are at least three ways to understand the question!

Q: Supongamos que vas a pasar 3 dias de un fin de semana de verano en un calido y maravilloso lugar – como el lago Garda (Norte de Italia), o la Playa Venice (Miami), o Pucon (Sur de Chile) – . Y digamos que quieres disfrutar – solo – 2 cockteles cada dia que estarüas alla. Y decides elegirlos entre los que son tus 7 cocteles favoritos, y no vas repetirlos. Puedes decir de cuantas maneras – distintas – podrias disfrutarlos??… Nota: Hay al menos tres maneras de interpretar la pregunta!

 

Note that there are at least three ways to understand the question! I actually wrote this quiz since I wanted to show:

  1. The most important & difficult thing in math usually is: UNDERSTANDING THE QUESTION.
  2. There always are different approaches to solve a problem.
  3. I found such a pretty formula!

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 4 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 4 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 4 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, the same number of ways of not choosing 1. 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 just

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 )

Source: Math-Update´s Blog

 

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