### Video Transcript

Find the possible values of π
which satisfy the equation 21Cπ equals 21C15.

Of course, we could take the
definition for combinations, which tell us πCπ equals π factorial over π
factorial times π minus π factorial. And then we could expand each of
these combinations to try and solve. However, sometimes we can solve
problems in a simpler and more straightforward manner by being familiar with
properties of combinations.

One such property is related to the
symmetry of combinations. This tells us that πCπ equals
πCπ minus π. If we apply that here, for our
first term weβll let π be equal to 21 and π be equal to π. And again for the right-hand side,
π equals 21, but 15 will be equal to π minus π to set up the symmetry of
combinations. If 15 equals π minus π and π
equals 21, we subtract 21 from both sides and see that negative six equals negative
π and that π equals six.

However, we should notice that
there is a possible solution as well where π equals 15, because of course 21C15
equals 21C15, meaning we have a possible solution at π equals 15 and a possible
solution at π equals six.