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

[eddy2017]

Hi, I am not understanding this problem very well
Out of 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 20.
Find the total number of elements that satisfy this condition.

I interpret this to mean that I have to find n, knowing that from a set of n elements you can choose 3 out of 20 different ways (regardless of the order, the order is not important). If this is the case, then I have to use the formula for combinations, right.?.
nCr = n! / r! * (n - r)!
But I am in doubt because I don't have the total of the elements,
thanks in advance for any tip/hint
eddy

 

[BigBeachBanana]

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

 

[eddy2017]

[math]n! / 3*2*1(n-3)!=20[/math]

 

[BigBeachBanana]

[math]n! / 3*2*1(n-3)!=20[/math]

Try to use parenthesis to indicate the correct order of operation. What about the n! and the (n-3)! Can you expand them?

 

[eddy2017]

[math](n!)/6!(n-3)!=20[/math][math](n!)/6*5*4*3*2*1(n-3)!=20[/math][math](n!)/720(n-3)!=20[/math]

 

[BigBeachBanana]

[math](n!)/6!=20[/math][math](n!)/6*5*4*3*2*1=20[/math][math](n!)/720=20[/math]

What happened to the (n-3)! Term?

 

[eddy2017]

we cross posted. check above

 

[eddy2017]

[math](n!)/6!(n-3)!=20[/math][math](n!)/6*5*4*3*2*1(n-3)!=20[/math][math](n!)/720(n-3)!=20[/math]

I am at a loss here. do I distribute the 720?

 

[BigBeachBanana]

How did 3*2*1 become 6!? Also you still haven’t attempt to expand n! and (n-3)! Term

 

[pka]

Out of 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 20.
Find the total number of elements that satisfy this condition.

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

[imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath]

 

[eddy2017]

How did 3*2*1 become 6!? Also you still haven’t attempt to expand n! and (n-3)!

[math](n!) / (3*2*1)(n-3)! =20[/math][math](n!) / 6(n-3)!=20[/math]

 

[eddy2017]

[math]6 (n!)/6(n-3)! = 6 (20)[/math][math]n!/(n-3)! =120[/math]

 

[eddy2017]

how do i solve that?

 

[Steven G]

EX: 24/4*3 = 6*3 NOT 6/3
24/(4*3) = 24/12 =2
So (n!)/(3∗2∗1)(n−3)! = [(n!)/(3∗2∗1)]*(n-3)!
You should write (n!)/[3*2*1*(n-3)!]
This was already pointed out by BigBeachBananas, but you chose to ignore his advice.

 

[eddy2017]

i put it in symbolab and i got this

 

[BigBeachBanana]

i put it in symbolab and i got this
View attachment 30350

You’re missing the 3! In the bottom

 

[eddy2017]

[math](n!) / (3*2*1)(n-3)! =20[/math][math](n!) / 6(n-3)!=20[/math]

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

 

[eddy2017]

You’re missing the 3! In the bottom

don't get it. what 3 in the bottom

 

[Steven G]

You were asked multiple times to simplify n!/(n-3)!
Just use the definition of factorial.

 

[eddy2017]

IF POST #17 is wrong then I think it is best to follow the formula given by pka.