There are multiple ways of going about this. We'll look at a few, using increasingly powerful combinatorial (counting) tools. Note that as a single pair is left over, this problem is equivalent to finding the number of ways to divide #8# distinct objects into #4# pairs.
Method 1: Direct Counting
Label the objects #1-8#. First, we'll pick the pair including #1#. There are #7# ways to do so: #(1,2), (1,3), ..., (1,8)#.
Next, we'll pick the pair including the least of the remaining six objects. There are #5# ways to do so. For example, if #(1,2)# was used, then they would be #(3,4), (3,5), ..., (3,8)#
Next, we'll pick the pair including the least of the remaining #4# objects. There are #3# ways to do so. The final pair will be the remaining #2# objects.
Putting it all together, that gives us #7*5*3 = 105# ways.
Method 2: Binomial Coefficients
The binomial coefficient #((n),(k)) = (n!)/(k!(n-k)!)# can be seen as the number of ways to choose #k# objects from a set of #n# objects. With that:
There are #((8),(2))# ways to choose #2# objects for the first pair. Then, there are #((6),(2))# ways to choose #2# from the remaining #6# for the second pair. Next, there are #((4),(2))# ways to choose #2# for the third pair, and then #((2),(2))# ways to choose the last pair.
Note, though, that this counts more than we wanted. Using the above, we are counting #(1,2),(3,4),(5,6),(7,8)# and #(3,4),(1,2),(5,6),(7,8)# as separate cases because it treats the first group as different from the second group and so on. To fix this, we can divide by the number of ways to arrange the groups. As there are #n!# ways to arrange a set of #n# objects, we'll divide by #4!# to account for different arrangements of the #4# pairs. Thus, the final result becomes
#((8),(2)) * ((6),(2)) * ((4),(2)) * ((2),(2)) * 1/(4!)#
#=(8!)/(2!cancel(6!)) * cancel(6!)/(2!cancel(4!)) * cancel(4!)/(2!cancel(2!)) * cancel(2!)/(2!0!) * 1/(4!)#
#=(8!)/(2!2!2!2!4!)#
#=105#
Method 3: Multinomial Coefficients
The multinomial coefficient #((,,n,,),(k_1, k_2,,..., k_m)) = (n!)/(k_1!k_2!...k_m!)# can be viewed as the number of ways to divide a set of #n# objects into #m# distinct groups of sizes #k_1# through #k_m#.
Applied to the given problem, we are dividing #8# objects into #4# groups, each of size #2#. Thus, we use the multinomial coefficient #((,,8,,),(2,2,2,2,))#. However, as this treats the groups as being distinct, we are overcounting, and must divide by the number of ways to arrange the pairs to account for repeated choices. As there are #4!# ways to arrange the #4# pairs, thus gives us the final result:
#((,,8,,),(2,2,2,2,))*1/(4!) = (8!)/(2!2!2!2!)*1/(4!) = 105#
