# "Allgemeine" Idee, um das Probieren einzuschränken für die Gleichung

#### polyglotwannabe

##### Senior Member
Hi, I need your help turning this phrase into understandable, sensible English
we're talking about a combinatorics exercise:

"Allgemeine" Idee, um das Probieren einzuschränken für die Gleichung.

Context

''Aus einer Gesamtzahl von Elementen werden 3er-Gruppen gebildet.

Unabhängig von der Reihenfolge, in der die Gruppen kombiniert werden

Die Gesamtzahl der Gruppen, die gebildet werden können, ist gleich 20.

Finden Sie die Gesamtzahl der Elemente, die diese Bedingung erfüllen:

(n/3)=20

"Allgemeine" Idee, um das Probieren einzuschränken für die Gleichung

(n/k)=m bei gegebenen positiven k,m:

(n/ k)=m bei gegebenen positiven k,m:

my translation is weak: General idea to restrict the intent of the equation..
Das Probieren is giving me trouble here.
thanks,
poly

• #### Thersites

##### Senior Member
Hello again Poly. I'm afraid I do not understand the context matter well enough to be of real help. But I can comment on
Das Probieren is giving me trouble here.
(Aus-)Probieren means giving something a try, having a shot at something and see if it works.

So here's my shot: General idea to limit the (necessary?) amount of trial and error in the process of establishing the (correct) equation.

I am sure that more competent help is already on the way. Have a good one, Thomas

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#### polyglotwannabe

##### Senior Member
that's it!. Trial and error. You hit the nail right on the head, Thomas!!. Thanks!. PASST!

#### Thersites

##### Senior Member
Just to be on the safe side you may want to wait for the comments of the others that will follow.

#### manfy

##### Senior Member
So here's my shot: General idea to limit the (necessary?) amount of trial and error in the process of establishing the (correct) equation.
If I understand the OP correctly then this is about right, but it sounds decidedly "unmathematical".
In fact the same is true for the German version.
In higher mathematics I'd expect to see something like:
"Allgemeiner Ansatz zur Reduktion der Iterationsschritte..." or similar.

There exist multiple approaches for solving equations.
Of course, for a simple algebraic equation like n/3=20, the obvious approach is the direct method: n=3*20=60.
But not all mathematical problems are solvable with direct methods. There are cases where empirical methods or iterative methods are preferrable and faster.
For example: You have a given set of potential solutions, the real numbers 1-100, and the mathematical problem with the relationship n/3=20. Find the correct solution in this given set of potential solutions by means of the empirical method.

So, now you can replace n with every single number in your given set and check the outcome for true or false.
If you start at 1, it will take 60 iterations before you get the valid solution 60.
The paragraph in the OP suggests that there are ways to limit that number of iterations by creating subsets of solutions that are limited to 3 elements. (As a side not, I doubt that this is a universally applicable method since it highly depends on the nature of the equation and the given set of solutions, but well, let's just look at it as a thought model.)

So, if that's what the OP means, I'd go along the lines of:
General approach for reducing/limiting iterations in empirical equation solving methods.

#### polyglotwannabe

##### Senior Member
Es geht um Binomialkoeffizienten Binomialkoeffizient – Wikipedia
„n über 3 gleich 20“

„Gruppe(n)“:
The correct German term is "Menge".
excatly!. and i made a mistake when i posted the exercise. it is not

it is nC3=20

#### anahiseri

##### Senior Member
Now the content is clear, a little comment related to grammar. The text looks like it consists of 4 sentences, as every line begins with a capital letter. But then the second sentence is not complete.
I think the best solution is to connect the second and the third, turning the second into a subordinate clause:

Unabhängig von der Reihenfolge, in der die Gruppen kombiniert werden,
ist die Gesamtzahl der Gruppen, die gebildet werden können, gleich 20.

#### polyglotwannabe

##### Senior Member
Now the content is clear, a little comment related to grammar. The text looks like it consists of 4 sentences, as every line begins with a capital letter. But then the second sentence is not complete.
I think the best solution is to connect the second and the third, turning the second into a subordinate clause:

Unabhängig von der Reihenfolge, in der die Gruppen kombiniert werden,
ist die Gesamtzahl der Gruppen, die gebildet werden können, gleich 20.
Yes, that is the way it should have been phrased. Thanks a lot.

#### manfy

##### Senior Member
Es geht um Binomialkoeffizienten Binomialkoeffizient – Wikipedia
„n über 3 gleich 20“
Aaaaaahh!!
I wouldn't have figured that one out in a 100 years with the German description in the OP!
Unabhängig von der Reihenfolge, in der die Gruppen kombiniert werden,
ist die Gesamtzahl der Gruppen, die gebildet werden können, gleich 20.
Yes, that's better but still far from clear.
If this is intended to give the student a hint for looking at the binomial coefficient (without explicitly mentioning that term), it would have to read:
Unabhängig von der Reihenfolge, in der die Elemente innerhalb der Gruppen kombiniert werden,​
ist die Gesamtzahl der Gruppen, die gebildet werden können, gleich 20.​
There is a fundamental difference between "Reihenfolge der Gruppen" and "Reihenfolge der Elemente in den Gruppen"!!

But even then, the title of this math problem, "Allgemeine" Idee, um das Probieren einzuschränken für die Gleichung", makes no sense whatsoever!
Where does "Probieren" come into the picture??

Once the student figures out that he can solve it with the formula for the binomial coefficient, he just solves the formula and gets n=6.

#### Schlabberlatz

##### Senior Member
Unabhängig von der Reihenfolge, in der die Elemente innerhalb der Gruppen kombiniert werden,
The correct term is "Menge". A "Gruppe" is something different: Gruppe (Mathematik) – Wikipedia

https://wikimedia.org/api/rest_v1/media/math/render/svg/c42a41f48e94296543f7f82ae26e19f69cc73ece
If I understand correctly, then it’s not necessary, strictly speaking, to mention that the ordering (?) should be ignored. You divide by k! in the binomial coefficient. You would have to leave out that division if the order were important. With n=6 and k=3 you would get nCr times 3! sets. That makes 120 sets instead of 20.

#### manfy

##### Senior Member
The correct term is "Menge". A "Gruppe" is something different: Gruppe (Mathematik) – Wikipedia
Touche! Here it was me who made the mistake of being imprecise.
https://wikimedia.org/api/rest_v1/media/math/render/svg/c42a41f48e94296543f7f82ae26e19f69cc73ece
If I understand correctly, then it’s not necessary, strictly speaking, to mention that the ordering (?) should be ignored. You divide by k! in the binomial coefficient. You would have to leave out that division if the order were important. With n=6 and k=3 you would get nCr times 3! sets. That makes 120 sets instead of 20.
No, actually it is necessary to define what is seen as a unique permutation.

If the sequence matters and (ABC) is formally different from (BAC), (CAB), etc. then n over k is solved with the formula n!/k! = 6!/3! = 120 n!/(n-k)! = 6!/(6-3)! = 120
But if the sequence does not matter and (ABC) = (BAC) you get the traditional formula of n!/(k!*(n-k)!) = 6!/3!*(6-3)! = 20.

PS: With the exception of the term 'binomial coefficient', not a single formula or math terminology I used in my posts was part of my active memory. I had to look it all up. All that stuff happened donkey's years ago and even though our math profs claimed that all this is crucially important knowledge, I didn't have to use it even a single time in all my professional career!

[edit: correcting my faulty factorial logic]

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#### Schlabberlatz

##### Senior Member
No, actually it is necessary to define what is seen as a unique permutation.
I don’t think so. As long as you don’t specify otherwise, a set is unordered. Ordered sets are called "Folge" or "Tupel".
Ist die Reihenfolge der Elemente von Bedeutung, dann spricht man von einer endlichen oder unendlichen Folge, wenn sich die Folgenglieder mit den natürlichen Zahlen aufzählen lassen (das erste, das zweite usw.). Endliche Folgen heißen auch Tupel.
Menge (Mathematik) – Wikipedia
Anyway, I said:
strictly speaking
It’s no harm if you say whether the order is important or not.

then n over k is solved with the formula n!/k! = 6!/3! = 120.
Nope. Your formula is wrong. It is not n!/k!, but n!/(n-k)!. It does not make a difference with n=6 and k=3, because 6-3=3. That’s a coincidence. Try it out with, say, n=6 and k=4.

And n!/(n-k)! is not called binomial coefficient. It is not nCr, but nCr times k!, see above.

#### manfy

##### Senior Member
Nope. Your formula is wrong. It is not n!/k!, but n!/(n-k)!. It does not make a difference with n=6 and k=3, because 6-3=3. That’s a coincidence. Try it out with, say, n=6 and k=4.

And n!/(n-k)! is not called binomial coefficient. It is not nCr, but nCr times k!, see above.
Thanks, you're right! You probably started replying before I had my correction posted.
Even though I didn't want to think about the whole thing anymore, it seems my subconscious kept mulling it over without me knowing and suddenly kept bugging me and I had to rethink it. I hate it when that happens and your own subconsious contradicts you.

#### polyglotwannabe

##### Senior Member
This is one simplest way our Math teacher defines binomial coefficient:

The Binomial Coefficients
The easiest way to explain what binomial coefficients are is to say that they count certain ways of grouping items. Specifically, the binomial coefficient C(n, k) counts the number of ways to form an unordered collection of k items chosen from a collection of n distinct items....
I know something about the topic, I'm not an expert thou. I just help with some translations that are successful because of you all.

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#### Hutschi

##### Senior Member
Gruppen zeigen Symmetrien. Es kann tatsächlich um Gruppen gehen.

PS:
Which part of math is it?
Groups are classified by symmetries.
Sets do not consider symmetry.

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#### polyglotwannabe

##### Senior Member
Gruppen zeigen Symmetrien. Es kann tatsächlich um Gruppen gehen.

PS:
Which part of math is it?
Groups are classified by symmetries.
Sets do not consider symmetry.

#### polyglotwannabe

##### Senior Member
I think it has to do with combinatorics.

#### Hutschi

##### Senior Member
General idea to limit the (necessary?) amount of trial and error in the process of establishing the (correct) equation.
... so I agree to the solution of Thomas with "trial and error".

#### polyglotwannabe

##### Senior Member
My first reaction was that too, cos I heard the math teacher mentioning that very same phrase.

#### Schlabberlatz

##### Senior Member
Es kann tatsächlich um Gruppen gehen.

Der Binomialkoeffizient ist also die Anzahl der k-elementigen Teilmengen einer n-elementigen Menge.
Binomialkoeffizient – Wikipedia

... so I agree to the solution of Thomas with "trial and error".
Ja, das dürfte passen. Man setzt die Werte in die Formel ein und erhält eine Gleichung. Man kann sie durch Ausprobieren lösen (eher nur bei kleinen Werten sinnvoll) oder man nimmt (bei Gleichungen höheren Grades) Polynomdivision. Da muss man auch ausprobieren, um einen passenden Wert zu finden. Man kann die Sache aber abkürzen (auch bei der erstgenannten Variante), wenn man bedenkt, dass eine mögliche Lösung Teiler des absoluten Gliedes sein muss.
"Allgemeine" Idee, um das Probieren einzuschränken für die Gleichung.

Das systematische Raten einer Lösung führt nur dann zum Erfolg, wenn es eine (leicht findbare) ganzzahlige Lösung gibt. Systematisch heißt in diesem Fall, dass wir unsere Suche auf die Teiler des absoluten Glieds beschränken. Der Zusammenhang zwischen Teiler des absoluten Glieds und Lösung der Gleichung folgt aus dem Satz von Vieta.
Kubische Gleichungen | Mathebibel

Bei k=3 und m=20 erhält man:
n³-3n²+2n-120=0.

Oder man multipliziert gar nicht aus und bleibt bei n(n-1)(n-2)=120.
Man probiert dann aus. n>3. Also kommen 4, 5, 6, 8, 10 etc. als Lösungen in Frage (Teiler von 120). 7, 9, etc. braucht man nicht zu überprüfen. Was aber hier unerheblich ist, denn die Lösung ist 6.

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#### Schlabberlatz

##### Senior Member
Ich würde nicht als erstes Fehler vermuten. Zumindest würde ich nachfragen, wenn ich Fehler vermute.
Die Sache war schon längst geklärt:
excatly!. and i made a mistake when i posted the exercise. it is not View attachment 59970

it is nC3=20
Wozu dann noch weiter nachfragen?

Es ist ja auch eine ganz konkrete Aufgabe mit konkreten Werten und einem eindeutigen Lösungsweg. nC3 [Edit: =20] steht für n über 3=20, also k=3, m=20, und man erhält als Ergebnis n=6, bumms, peng, Ende, aus. Gruppen kommen nicht vor.

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#### polyglotwannabe

##### Senior Member

Ja, das dürfte passen. Man setzt die Werte in die Formel ein und erhält eine Gleichung. Man kann sie durch Ausprobieren lösen (eher nur bei kleinen Werten sinnvoll) oder man nimmt (bei Gleichungen höheren Grades) Polynomdivision. Da muss man auch Ausprobieren, um einen passenden Wert zu finden. Man kann die Sache aber abkürzen (auch bei der erstgenannten Variante), wenn man bedenkt, dass eine mögliche Lösung Teiler des absoluten Gliedes sein muss.

Bei k=3 und m=20 erhält man:
n³-3n²+2n-120=0.

Oder man multipliziert gar nicht aus und bleibt bei n(n-1)(n-2)=120.
Man probiert dann aus. n>3. Also kommen 4, 5, 6, 8, 10 etc. als Lösungen in Frage (Teiler von 120). 7, 9, etc. braucht man nicht zu überprüfen. Was aber hier unerheblich ist, denn die Lösung ist 6.

#### polyglotwannabe

##### Senior Member
Thanks to you all. What I like most about you all is the passion that you put into your posts and answers. you really care!. Amazing!.