There are 8 people in a room. They were asked to go in a certain place 2 at a time. In how many ways could the people be paired? Is it an example of

Thanks Alan I do believe you are correct :) 

( My first answer was rubbish )

I have more ideas:

There are 8 people in a room. They were asked to go in a certain place 2 at a time. In how many ways could the people be paired? Is it an example of permutation or combination problem?

I'll pair them first   [ This is a combination question because order doesn't matter]

choose anyone then there are 7 ways to choose a mate for that person

choose anyone then there are 5 people left to pair him with

choose anyone then there are 3 people left to pair him with

there there is only 2 left so they have to go together

so that is 7*5*3 = 105 possible pairs. [ Just like Alan said ]

Now they leave the room in pairs and order counts so that would be 4! = 2*3*4 = 24   [This bit is a permutation question]

So now I get     7*5*3*4!         105*24 = 2520

So yes I agree that there is 105 ways for them to be paired ( and I think that is what the answer was asking)

 but I think there is 24 orders that those pairs can leave the room.  So there are 105*24 = 2520  ways they can leave the room 

Dunno and don't care

@  Injustagod, if you speak to people that way you may find the help you want for yourself stops being offered !

There are 8 people in a room. They were asked to go in a certain place 2 at a time. In how many ways could the people be paired? Is it an example of permutation or combination problem?

Deleted on the grounds that my answer was rubbish   

I think there are 105 ways of pairing: (7*5*3 = 105)

The figure below illustrates the pattern (though I haven't filled it in completely)

Consider this the same as unique combinations of 2 from integer set {n}

1 person, 0 pair

2 persons, 1 Pair
3 persons, 3 Pairs
4 persons, 6 Pairs
5 persons, 10 Pairs
6 persons, 15 Pairs

7 persons, 21 Pairs

8 persons, 28 Pairs

Note the consistent relationship to

\(\text {unique pairs = } \large \frac{n(n-1)}{2} \scriptsize \text { for n} \geq 0\\ \)

Or

\(\text {unique pairs = } \Large \binom {n}{2} \scriptsize \text { for n} \geq 2\\ \)

--------

Hi Nauseated,

Your answer:

Consider this the same as unique combinations of 2 from integer set {n}

1 person, 0 pair

2 persons, 1 Pair
3 persons, 3 Pairs
4 persons, 6 Pairs
5 persons, 10 Pairs
6 persons, 15 Pairs

7 persons, 21 Pairs

8 persons, 28 Pairs

nC2 unique pairs   

-----------------------------------------------------------------------------------

You have found how many individual pairs can be made from 8 people.

That is NOT the question.

If there is three people, or any odd number of people in a room, they cannot all be paired of so the question would make no sense.  There would be no answer.

 You have said that if there are 4 people in a room there would be 6 pairs. This is not correct.

These are the the possible combinations

AB    and   CD

AC   and   BD

AD   and   BC

There are no other possible combinations.  So there are 3 combinations.

If there are 6 people in a room then you are correct, there are 15 combination.

because

IF E and F are together there are 3 ways the others can be combined, as shown above.

IF E and A are together there are 3 ways the others can be combined, as shown above.

IF E and B are together there are 3 ways the others can be combined, as shown above.

IF E and C are together there are 3 ways the others can be combined, as shown above.

IF E and D are together there are 3 ways the others can be combined, as shown above.

So there is a total of 3*5 = 15 ways that 6 people can be paired off

Now there  there is 8 people in a room    A, B, C, D, E, F, G, H

G and H are together then there are 15 ways the others can be paired.

G and A are together then there are 15 ways the others can be paired.

G and B are together then there are 15 ways the others can be paired.

G and C are together then there are 15 ways the others can be paired.

G and D are together then there are 15 ways the others can be paired.

G and E are together then there are 15 ways the others can be paired.

G and F are together then there are 15 ways the others can be paired.

G and  are together then there are 15 ways the others can be paired.

So there are  15*7 = 105 ways that the people can be paired

etc

If there are 2n people in the room then there will be 1*3*5*7*........*(2n-1)  ways that they can be paired.

According to OEIS   (Online Encyclopaedia of Integer Sequences)

https://oeis.org/search?q=1%2C3%2C15%2C105%2C&language=english&go=Search

The notation for this is   (2n-1)!!

You have found how many individual pairs can be made from 8 people.

That is NOT the question.

----

OK, I may have misinterpreted the question. You explained yours and Alan’s focus clearly.

My focus is on this part: “. . . In how many ways could the people be paired?” This is where the question mark is. However, if the sequence of these unique pairs entering a room is a consideration then my solution is obviously incomplete. 

-------

If there is three people, or any odd number of people in a room, they cannot all be paired of so the question would make no sense.  There would be no answer.

---------

Actually, they can, just not all of them simultaneously when the count is odd. 

Reiteration: This formula gives the unique combinations of pairs (sets of 2) from a set of {n}. 

It’s rather interesting why this works for both odd and even numbers. This sequence is an offset of Gauss’ “Sum of Sequential Numbers” formula. However, in this case, the formula adds, not the first N numbers, but the first (N-1) numbers, aligning it with the unique combinations of 2 from set {n}. (This does not have a unique OEIS ID). 

Here's why even and odd counts solutions use the same formula in Gauss’ “Sum of Sequential Numbers.”

For an even count:

\( \text{Number of pairs * Sum of each pair} \large = (\frac{n}{2})(n+1) = \frac{n(n+1)}{2}\)

For an odd count

\(\text{Number of pairs * Sum of each pair} = \Large (\frac{n + 1}{2})(n) = \frac{n(n+1)}{2} \)