Official solution
(
19Cp)max =
19C9 or
19C10 = a
(
20Cp)max =
20C10 = b
(
21Cr)max =
21C10 or
21C11 = c
1 = \[{a \over {^{19}{C_{10}}}} = {b \over {^{20}{C_{10}}}} = {c \over {^{21}{C_{10}}}}\]
\[ \Rightarrow \] \[{a \over {11}} = {b \over {22}} = {c \over {42}}\]