Need help in dividing a set into different pairs

Question

In how many ways can you divide the set of eight numbers {2,3,4,5,6,7,8,9} into 4 pairs such that no pair of numbers has G.C.D equal to 2?

Answer 1

Given this set of numbers, the only way for a pair of them to have a greatest common denominator of $2$ would be for both of them to be even... with one exception. We are allowed to pair $(4, 8)$, because in this case they will have a GCD of $4$, not $2$.

So let's split this problem into (a) the cases where $(4, 8)$ are not a pair, and (b) the cases where $(4, 8)$ are a pair.

In case (a), we have to pair each even number with an odd number. So our pairs are $(2, \_)$, $(4, \_)$, $(6, \_)$, and $(8, \_)$. We have to place the four remaining odd numbers $\{3, 5, 7, 9\}$ into these blanks. So there are $4! = 24$ total possibilities here.

In case (b), we know that $(4, 8)$ are a pair. However we still aren't allowed to make the pair $(2, 6)$, as these would have a GCD of $2$. But other than this, anything is game. The remaining numbers are $\{2, 3, 5, 6, 7, 9\}$. Let's think of the pairs this time as $(2, \_)$, $(6, \_)$, and $(\_, \_)$ There are $4$ options for the first blank, $3$ options for the second blank, and after this the last two are determined. This makes $4\cdot 3 = 12$ possibilities here.

So all in all, I count $24+12=36$ total possible ways.

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EDIT: you can also verify that this reasoning is correct by exhaustively listing all 36 possible pairings. Here are the ones I came up with:

1:  (2, 3) (4, 5) (6, 7) (8, 9)
2:  (2, 3) (4, 5) (6, 9) (7, 8)
3:  (2, 3) (4, 7) (5, 6) (8, 9)
4:  (2, 3) (4, 7) (5, 8) (6, 9)
5:  (2, 3) (4, 8) (5, 6) (7, 9)
6:  (2, 3) (4, 8) (5, 7) (6, 9)
7:  (2, 3) (4, 8) (5, 9) (6, 7)
8:  (2, 3) (4, 9) (5, 6) (7, 8)
9:  (2, 3) (4, 9) (5, 8) (6, 7)
10: (2, 5) (3, 4) (6, 7) (8, 9)
11: (2, 5) (3, 4) (6, 9) (7, 8)
12: (2, 5) (3, 6) (4, 7) (8, 9)
13: (2, 5) (3, 6) (4, 8) (7, 9)
14: (2, 5) (3, 6) (4, 9) (7, 8)
15: (2, 5) (3, 7) (4, 8) (6, 9)
16: (2, 5) (3, 8) (4, 7) (6, 9)
17: (2, 5) (3, 8) (4, 9) (6, 7)
18: (2, 5) (3, 9) (4, 8) (6, 7)
19: (2, 7) (3, 4) (5, 6) (8, 9)
20: (2, 7) (3, 4) (5, 8) (6, 9)
21: (2, 7) (3, 5) (4, 8) (6, 9)
22: (2, 7) (3, 6) (4, 5) (8, 9)
23: (2, 7) (3, 6) (4, 8) (5, 9)
24: (2, 7) (3, 6) (4, 9) (5, 8)
25: (2, 7) (3, 8) (4, 5) (6, 9)
26: (2, 7) (3, 8) (4, 9) (5, 6)
27: (2, 7) (3, 9) (4, 8) (5, 6)
28: (2, 9) (3, 4) (5, 6) (7, 8)
29: (2, 9) (3, 4) (5, 8) (6, 7)
30: (2, 9) (3, 5) (4, 8) (6, 7)
31: (2, 9) (3, 6) (4, 5) (7, 8)
32: (2, 9) (3, 6) (4, 7) (5, 8)
33: (2, 9) (3, 6) (4, 8) (5, 7)
34: (2, 9) (3, 7) (4, 8) (5, 6)
35: (2, 9) (3, 8) (4, 5) (6, 7)
36: (2, 9) (3, 8) (4, 7) (5, 6)

Answer 2

$2$ must be paired with an odd number.

$6$ can not be paired with an even number that doesn't also have three as a divisor so $6$ must be paired with an odd number.

$4$ may (but doesn't have to be) paired with $8$ and there are no other restrictions.

There are $4$ odd numbers to pair with TWO.

And after that there are $3$ remaining odd numbers to pair with SIX.

And for the remaining four numbers there are ${4\choose 2}$ ways to choose a the third pair and ${2 \choose 2} =1$ way to choose the last pair.

But choosing $(a,b)$ then $(c,d)$ is the same as $(c,d) $ then $(a,b)$ so this is double counting. There are $\frac 62 = 3$ ways to choose the final two pairs.

(Alternatively given any of the $4$ remaining unpaired numbers there are $3$ numbers we can pair with it. And the remaining to numbers must pair together.)

So there are $4*3*3 = 36$ such ways.