Originally Posted by

**Plato** I just don't see it. I do not think you showed that. Please explain.

$\begin{array}{*{20}{c}} \text{left}&\| & {{L_1}}&{{L_2}}&{{L_3}}&{{L_4}}&{{L_5}}&{{L_6}} \\ \text{right}&\| & {{R_1}}&{{R_2}}&{{R_3}}&{{R_4}}&{{R_5}}&{{R_6}} \end{array}$ there are $\dbinom{12}{4}=495$ ways to select a four element subset from that twelve element set.

Suppose we choose the subset $\{L_1,L_2,L_3,R_4\}$ has no two elements have the same subscript.

Thus we have not selected a pair. How many four element subsets **do not have two elements** with the same subscript?

Some of those contain all $L_k's$; some all $R_k's$ ; some contain three $L_k's$ and one $R_k$; some contain three $R_k's$ and one $L_k$; AND some contain two of each. Now some of those are counted the same way. The first two and the next two the same number of ways.

Subtract that from the total and you have the number of subsets of four that contain at least one pair.