Find the total number of elements that satisfy this condition: (n/3)=20

[eddy2017]

hi to all teachers;
I'm having trouble understanding the end of this problem.
if you don't want to help is okay, i am just posting it. if you do help, please follow the 'I'M STUCK' ,WHAT NOW? , MAKE IT AS EASY AS POSSIBLE. MATH IS NOT SO COMPLICATED IF EXPLAINED CLEARLY mantra.

This is it:

From a total of elements, groups of 3 are made.
Regardless of the order in which the groups are combined.

The total number of groups that can be formed is equal to 20.

Find the total number of elements that satisfy this condition:

(n/3)=20

I have seen this exercise done on a video and I understood up to a certain point.

The formula for combinations is

(n/k)=n!/((n-k)!k!)= 20

We have that k =3 ( groups of 3)

I substitute k for its value,

Then,

n/3= n!/(n-3)!3! =20

the teacher, in the video, decides to clear n!

multiplying (n-3)! 3! on both sides of the equality,

Thus, the result is,

n!=20(n-3)! 3!

then he says:

To construct a factorial number you know that you must have all the integer factors in an increasing way and none missing.

We go on to decompose the factorials ...

I understood this

20 (n-3)! =? 3! = 3*2*1

4*5

2*4

factors =1,2,3,3,4,5

then,

n!=1*2*3*4*5(n-3)

Then says the teacher,

I quote 'so far you have that n! is greater than 5! And he points out with an arrow all the factorial of 5 and if we want to continue constructing the factorial number we have to have a 6...
i lost him there. I did not understood what he did after that.
the video was in Spanish but i think i have translated everything to the letter. i did not miss anything up to the point where i stopped understanding.

thanks for any tip that leads to a solution,
eddy

 

[pka]

Sorry to say, what you posted above makes no sense whatsoever.
If you will post a link to the video I think that one of us can gladly help you.

 

[eddy2017]

Sorry to say, what you posted above makes no sense whatsoever.
If you will post a link to the video I think that one of us can gladly help you.

From a total of elements, groups of 3 are made.
Regardless of the order in which the groups are combined. The total number of groups that can be formed is equal to 20.
Find the total number of elements that satisfy this condition:

(n/3)=20

Okay thanks, pka
This is the link.
It is in Spanish but maybe by seeing the operation you get an idea.

 

[eddy2017]

can't be played.

 

[eddy2017]

I don't think it is working
It says it is unavailable when I click on it.

 

[eddy2017]

The formula for combinations is given

(n/k)=n!/((n-k)!k!)= 20
We have that k =3 ( groups of 3)

I substitute that for its value

Then,

n/3= n!/(n-3)!3! =20
the teacher, in the video, decides to isolate n!

multiplying (n-3)! 3! on both sides of the equality,

Thus, the result is,

n!=20(n-3)! 3!

Now he says:

To construct a factorial number you know that you must have all the integer factors in an increasing way and none missing.

We go on to decompose the factorials ...

20 3! in factors= 3*2*1

4*5

2*4

we dont know the value of n yet

3! in factors= 3*2*1

factors =1,2,3,3,4,5

then,

n!=1*2*3*4*5(n-3)

Then he says

and I quote 'so far it is found that n! is greater than 5! And he points out with an arrow all the factorial of 5 and if we want to continue building the factorial we have to have a 6...

This is really where I got lost, I no longer understood the reason for the other ones.
don't know if this makes any sense to you

 

[Otis]

My approach is different.

We know that 3! is 6, so I would multiply both sides by 6 right away.

n!/(n-3)! = 120

The left-hand side may be simplified.

Let's look at an example, to see how it simplifies. Let n = 10.

10!/(10-3)! means (1*2*3*4*5*6*7*8*9*10)/(1*2*3*4*5*6*7)

After cancelling common factors, we get 8*9*10

Think about 8*9*10 with respect to symbol n. We can express 8*9*10 in terms of n. Can you see how?

Spoiler: Answer

The simplification is (n)(n-1)(n-2)

If you're not sure why, then write out more examples and cancel common factors until it becomes clear. Let n=8. Let n=9. Let n=12…

After simplifying the left-hand side, we can solve the equation by inspection.

Spoiler: Hint

Determine the prime factorization of 120, and go from there, remembering that we're looking for three consecutive Integers whose product is 120.

?

 

[eddy2017]

okay, good. thanks a lot, Otis.
I will try to do it using your approach. i did not fully understand the other.
one question:
does this have to do with calculation of binomial coefficients?

 

[eddy2017]

a german friend sent me this approach but i do not know where he is getting the 1 and the 2.


and this what he says translated into english
Yes, you could take (n3)=n(n-1)(n-2)/6=20
as the equation of determination for n and solve it (as a cubic equation).

But you can also try n=3,4,... until it works - probably faster .

 

[Otis]

does this have to do with calculation of binomial coefficients?

I'm not sure what you're referring to, when you say "this".

If you're asking whether this thread has to do with calculating binomial coefficients, then the answer is no.

However, when we do expand a binomial power like (x+1)^n, then we use n!/[k!(n-k)!] to calculate the resulting polynomial's coefficient for each x^k term.

Check it out..

?

 

[Otis]

i do not know where he is getting the 1 and the 2

Work through the suggestions that I provided to you, in post #7.

If that's too confusing, then please post why.

If n is an Integer, how do we use symbol n to express the two Integers that come before n?

?

 

[eddy2017]

Thanks, very good

 

[eddy2017]

Work through the suggestions that I provided to you, in post #7.

If that's too confusing, then please post why.

If n is an Integer, how do we use symbol n to express the two Integers that come before n?

?

I'll do that. Tomorrow I will explore it. Now it is too late for me. Just letting you know cos don't want you to think I have given up on this.
If I get stuck tomorrow, I'll reach out.
Thanks a lot for support.
One question:
What standard would you tie this this type of problem/ exercise to if you had to assign it a standard/skill?

 

[Otis]

This is a combinatorics exercise. The subject begins with learning about factorial expressions and how to simplify them. Next we learn how to count "Combinations" and "Permutations".

?

 

[eddy2017]

Thank you!.

 

[eddy2017]

My approach is different.

We know that 3! is 6, so I would multiply both sides by 6 right away.

n!/(n-3)! = 120

The left-hand side may be simplified.

Let's look at an example, to see how it simplifies. Let n = 10.

10!/(10-3)! means (1*2*3*4*5*6*7*8*9*10)/(1*2*3*4*5*6*7)

After cancelling common factors, we get 8*9*10

Think about 8*9*10 with respect to symbol n. We can express 8*9*10 in terms of n. Can you see how?

Spoiler: Answer

The simplification is (n)(n-1)(n-2)

If you're not sure why, then write out more examples and cancel common factors until it becomes clear. Let n=8. Let n=9. Let n=12…

After simplifying the left-hand side, we can solve the equation by inspection.

Spoiler: Hint

Determine the prime factorization of 120, and go from there, remembering that we're looking for three consecutive Integers whose product is 120.

?

hey, hi, hope you're doing well.
this is how i developed the exercise so far. i will need your guidance when i get stuck. i will ask you. be as direct as possible.
let me solve it this way and then i can try your way. i wanna try your way, but i'm not getting it. let me try the one I'm most familiar with but i do not understand completely either.

if there is any sense in what i am doing, even if it is not the approach you would choose, then let me go till the end just prop me up when i falter

so the problem is asking me to find the numbers that fulfill this condition:

n/3= 20

i will apply the combination formula

nCr= n! /(n-r)! r!

where n=total of things in the set
r= number of choosing objects from the set.

and then the problem equates this setup to 20

nCr= n! /(n-r)! r!= 20

now, let's see the values i have
i have that r=3
now, i'll plug it into the formula

nCr= n! /(n-r)! r!= 20

nCr= n! /(n-3)! 3!= 20

okay, look now, this method calls for isolating n!
so i'll divide (n-3)! r! into both sides to clear away the fraction,

we're left with
n! =20 (n-3)! 3!
now i am going to try and factor these three terms: 20, (n-3)!, and 3!

20 = 4*5
4=2*2

(n-3)! don't know its value yet.

3! = 3*2*1

now organizing all factors
1*2*3*4*5 * (n-3)!


i stopped understanding the video right here, otis.
if you want to help me please, please, help me finish this up.
if you want to ask me what comes after that, let me know, and i will tell you what was that i did not understand after this.
then we can pursue it another way.
this video is form a guy who has a website and teaches math so i do not think this is wrong. might not be the best way, but i don't think it is wrong. help finish where i left off . this is the visual that i have. i am a visual learner.
thank you for caring and trying to help.

 

[Dr.Peterson]

so the problem is asking me to find the numbers that fulfill this condition:

n/3= 20

You really need to type this notation correctly. What you wrote says that n divided by 3 is 20; the answer to that is that n is 60. You've written this before as (n/3) and as (n 3), neither of which means what you mean.

What you mean is [imath]{n\choose 3}=20[/imath], which is something very different. There are several ways you can write it without needed LaTeX. One that many students use is nC3; another that I prefer is C(n,3).

How to solve it is a separate matter; but you will get better help in general when you make it easy to figure out what you are asking.

does this have to do with calculation of binomial coefficients?

Yes, it does, though the problem itself is not about calculating them, but the reverse. This is another term for "combinations"; that's why an alternative notation is nC3, or properly [imath]_nC_3[/imath].

 

[eddy2017]

Thank you Dr Peterson, indeed!. Yes, that one is the exact notation!

 

[eddy2017]

Thank you Dr Peterson, indeed!. Yes, that one is the exact notation!

Could you recommend a video where they deal with this type of exercise, a video where they solve it explaining steps???.

 

[Dr.Peterson]

Could you recommend a video where they deal with this type of exercise, a video where they solve it explaining steps???.

As far as I'm concerned, this is not a procedure that can or should be taught; it is a non-standard problem that makes you think. So I wouldn't look for a video about it.

Here is how I would approach it, putting together other things that have been said:

The simpler way to think of [imath]{n\choose 3}[/imath] is as [imath]\frac{n(n-1)(n-2)}{(3)(2)(1)}[/imath]; that is, the product of 3 numbers descending from n, over the product of 3 numbers descending from 3.

So you have the equation [imath]\frac{n(n-1)(n-2)}{6}=20[/imath], or [imath]n(n-1)(n-2) = 120[/imath].

You have found that 120 factors as 2*3*4*5, or in terms of prime factors, [imath]2^3\cdot 3\cdot 5[/imath]. From here, the rest is "by inspection", which really means you try to arrange the factors into three consecutive numbers. This part can't be taught; you just have to try. But I'll give you a hint: what is 2*3?