If a, b and c are the greatest values of 19Cp, 20Cq and 21Cr respectively, then:
Solutions · 1
Official solution
nCr is greatest at the middle term(s): a=19C9=19C10, b=20C10, c=21C10=21C11.
\[\frac{a}{1} : \frac{b}{2\cdot\frac{20}{10}\cdot\ldots}\]
Using ratios ab=19C920C10=2, and ac=19C921C11=1142:
\[\frac{a}{11}=\frac{b}{22}=\frac{c}{42}\]