Find the total number of elements that satisfy this condition..

[Steven G]

don't get it. what 3 in the bottom

You are amazing!
BBB says that you did not put 3! in the denominator and you respond with what 3.

3! and 3 are different.
3!=6. So what happened to the 6 in the denominator.

 

[Steven G]

$$n!/(3*2*1)(n-3)! =[(n!)/(3*2*1)]*(n-3)!$$(n!)/[3*2*1*(n-3)!$$$$(n!)/6(n-3)!$$$$ how about now$$

Still not correct. You opened a bracket, [ , but you never closed it!

 

[eddy2017]

You are correct that you have to find n. What do you know about factorial? Can you expand them? What do you notice?
$$\frac{\text{n!}}{3!(n-3)!}=20$$

this was BBB original equation. I will repeat calculations

 

[eddy2017]

$$n!/3*2*1(n-3)! =20$$n!/6(n-3) =20

 

[eddy2017]

is this correct or not?. BBB set it up at post #2

 

[eddy2017]

$$or following jomo's suggestion$$$$[n!)/3*2*1*(n-3)!]$$

 

[BigBeachBanana]

$$n!/3*2*1(n-3)! =20$$n!/6(n-3) =20

This equation is not equivalent to what I set up in post#2 because the lack of parenthesis.

 

[eddy2017]

$$n!/6(n-3)!=120$$

 

[eddy2017]

is that setup correct or not

 

[eddy2017]

You need to learn this notation. We have N things choosing three at a time.
$^N\mathcal{C}_3=\dbinom{N}{3}=\dfrac{N!}{(3!)(N-3)!}=20$
SEE HERE

$$$$$$$$$$

yes, but I want to see how we get to the result of 6. I need to work the formula myself. I need to do it so I understand the steps

 

[eddy2017]

$$n!/6(n-3)!=120$$

Let's focus on what to do next from here.

 

[eddy2017]

Drop me a hint

 

[pka]

O.K. here it is n!/(n−3)!=120 is $n^3-3n^2+2n=120$
OR $n^3-3n^2+2n-120=0$ Now I have no idea of how to deal with that cubic.
HERE is how I would find out that $n=6$.

 

[eddy2017]

Therecis no other way except the quadratic way?. Because it is really neat but I do not get how you get to the n^3.......

 

[BigBeachBanana]

$$n!/(3*2*1)(n-3)! =[(n!)/(3*2*1)]*(n-3)!$$(n!)/[3*2*1*(n-3)!$$$$(n!)/6(n-3)!$$$$ how about now$$

What you've written is here is $$\frac{n!(n-3)!}{3!}$$which is not the same as
$$\frac{\text{n!}}{3!(n-3)!}$$

Drop me a hint

I've already given you a hint from post#2. I asked you to use the definition of factorial to simplify the n! and the (n-3)! terms.

 

[eddy2017]

BBB, I do not know how to expand the factorial of a variable. All I can do is this. If you help me I can continue.
n!/3*2*1(n-3)!=20

 

[pka]

Therecis no other way except the quadratic way?. Because it is really neat but I do not get how you get to the n^3.......

If you cannot multiply $n(n-1)(n-2)$ and get $n(n^2-3n+2)=n^3+3n^2+2n$$$$$
The you are wasting all our time here!

 

[BigBeachBanana]

BBB, I do not know how to expand the factorial of a variable. All I can do is this. If you help me I can continue.
n!/3*2*1(n-3)!=20

You knew 3! = 3* 2*1 = 6
So if I ask you what's n!? What would you do? Use the definition of factorial. It is the product of all positive integers less than or equal to a given positive integer and denoted by that integer and an exclamation point.

$$\frac{\text{n!}}{3!(n-3)!}=\frac{n(n-1)(n-2)(n-3)...(3)(2)(1)}{6(n-3)(n-4)(n-5)...(3)(2)(1)}=20$$Now, what do you notice between the numerator and denominator?

 

[eddy2017]

I see that (3) (2) (1) are in both numerator and denominator and that they can cancel out

 

[BigBeachBanana]

I see that (3) (2) (1) are in both numerator and denominator and that they can cancel out

Ok, what about the other terms?