If 𝑛𝐶42 plus 𝑛𝐶𝑛 minus 42 equals
two times 𝑛𝐶43, find 𝑛.
Looking at this equation, we have
three combinations, all of which have a set size of 𝑛, but a different size 𝑟. Recall that the definition for the
number of combinations of size 𝑟 taken from a set size 𝑛 is given by 𝑛 factorial
over 𝑟 factorial times 𝑛 minus 𝑟 factorial.
Now, one strategy for solving here
would be to write each of these combinations in this expanded definition form. However, after we examine the two
combinations on the left, we recall a property that we can use to simplify. And that’s the symmetry property of
combinations. It tells us that 𝑛𝐶𝑟 equals 𝑛𝐶𝑛
minus 𝑟. That is, the combination size 𝑟
from set 𝑛 will be equal to the combination of size 𝑛 minus 𝑟 of a size set
𝑛. By this property, both of the
combinations on the left side of this equation are equal to each other.
This means we could rewrite our
first term 𝑛𝐶42 as 𝑛𝐶𝑛 minus 42. After we bring everything else
down, we can simplify to two times 𝑛𝐶𝑛 minus 42 equals two times 𝑛𝐶43. The twos on both sides cancel
out. This implies that 𝑛 minus 42 would
be equal to 43. And if 𝑛 minus 42 equals 43, then
𝑛 would equal 85.
Because we know of the symmetry
property, it’s also worth checking the second solution that could be possible. For 𝑛𝐶𝑎 equal to 𝑛𝐶𝑏, the
options are 𝑎 equals 𝑏, which we found here, or 𝑎 equals 𝑛 minus 𝑏. That second solution would be 𝑛
minus 42 equals 𝑛 minus 43, which is not consistent. This means there’s only one
solution for 𝑛 in this combination equation. If you wanted to check that this
was correct, you could plug back in 85 for 𝑛 and solve the combinations on your