Possible combinations

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ssmmss

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If we have 100 people, how may combinations of 5 people teams are possible. I'm thinking this is a factorial:
=100*99*98*97*96.

On a different note, I'm starting to think it's 5^100.
 
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ssmmss said:
If we have 100 people, how may combinations of 5 people teams are possible. I'm thinking this is a factorial:
=100*99*98*97*96.
Click to expand...
The way you have written this, there will be twenty teams in all. \(\displaystyle \dfrac{100!}{(5!)^{20}(20!)}\)

OR did you mean one team of five persons?
 
pka said:
The way you have written this, there will be twenty teams in all. \(\displaystyle \dfrac{100!}{(5!)^{20}(20!)}\)

OR did you mean one team of five persons?
Click to expand...

What I meant was how many different teams of 5 are possible out of the 100.
 
ssmmss said:
What I meant was how many different teams of 5 are possible out of the 100.
Click to expand...

You still do not understand the language.
The way you have written this it means:
"How many ways can 100 people be grouped into teams of five each?"

But that is not the same as:
"How many ways can a team of 5 be chosen from 100 people?"

Do you see the difference?

Which of those do you mean?
 
pka said:
You still do not understand the language.
The way you have written this it means:
"How many ways can 100 people be grouped into teams of five each?"

But that is not the same as:
"How many ways can a team of 5 be chosen from 100 people?"

Do you see the difference?

Which of those do you mean?
Click to expand...

Thanks for pointing that out. Definitely the 2nd scenario: "How many ways can a team of 5 be chosen from 100 people?"
 
ssmmss said:
Thanks for pointing that out. Definitely the 2nd scenario: "How many ways can a team of 5 be chosen from 100 people?"
Click to expand...

Then the answer is \(\displaystyle \dbinom{100}{5}=\dfrac{100!}{(5!)(95!)}\)
 
pka said:
Then the answer is \(\displaystyle \dbinom{100}{5}=\dfrac{100!}{(5!)(95!)}\)
Click to expand...

Thank you pka, I see you used the definition of k combinations where the order doesn't matter and the elements can't be repeated, which makes sense.
 
So if we take this scenario further and we say:
The 100 people are comprised of 20 women and 80 men and the team must contain exactly 2 women, then how many different 5 person teams are possible? (order doesn't matter).

Then, I would do: 20!/2!*18!=190
80!/3!*77!=82160

answer is: 190+82160
 
ssmmss said:
So if we take this scenario further and we say:
The 100 people are comprised of 20 women and 80 men and the team must contain exactly 2 women, then how many different 5 person teams are possible? (order doesn't matter).
Then, I would do: 20!/2!*18!=190
80!/3!*77!=82160
answer is: 190+82160
Click to expand...
Multiply not add!
\(\displaystyle \dbinom{80}{3}\cdot\dbinom{20}{2}\)
 
Denis said:
In case interested; a,b = 2 women, c,d,e = 3 men
Looper that "gets 'em all:

FOR a = 1 to 19
FOR b = a+1 to 20
FOR c = 21 to 98
FOR d = c+1 to 99
FOR e = d+1 to 100

With 15,610,400 "teams", you'd need quite a bit of paper to print 'em all :p
Click to expand...

Nice explanation. Thanks Denis.
 
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