If a, b and c are the greatest values of $$^{19}C_p$$, $$^{20}C_q$$ and $$^{21}C_r$$ respectively, then:

For every binomial coefficient we recall the symmetry and monotonicity property

$$^nC_r = \; ^nC_{n-r}$$

and, starting from $$r = 0$$, the values of $$^nC_r$$ keep on increasing as $$r$$ increases up to the integral part of $$\dfrac{n}{2}$$, after which they start decreasing. This means that the greatest term (or terms) of the expansion of $$(1+x)^n$$ occur(s) at

$$r = \left\lfloor \dfrac{n}{2} \right\rfloor \quad\text{or}\quad r = \left\lceil \dfrac{n}{2} \right\rceil.$$

Now we apply this fact one by one.

For $$^{19}C_p$$ we have $$n = 19$$. The integral part of $$\dfrac{19}{2}$$ is $$9$$ and the next integer is $$10$$. Hence the greatest value is attained at $$p = 9$$ or $$p = 10$$, and both values are the same:

$$a = \;^{19}C_9 = \;^{19}C_{10}.$$

Using the factorial definition,

$$a = ^{19}C_9 = \dfrac{19!}{9!\,10!}.$$

Simplifying step by step,

$$\begin{aligned} a &= \dfrac{19\cdot18\cdot17\cdot16\cdot15\cdot14\cdot13\cdot12\cdot11}{9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1} \\ &= 92378. \end{aligned}$$

For $$^{20}C_q$$ we have $$n = 20$$. Now $$\dfrac{20}{2}=10$$, so the single greatest value is at $$q = 10$$.

$$b = \;^{20}C_{10} = \dfrac{20!}{10!\,10!}.$$

Again expanding only what is needed,

$$\begin{aligned} b &= \dfrac{20\cdot19\cdot18\cdot17\cdot16\cdot15\cdot14\cdot13\cdot12\cdot11}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1} \\ &= 184756. \end{aligned}$$

For $$^{21}C_r$$ we take $$n = 21$$. Now $$\dfrac{21}{2}=10.5$$, so the two central positions $$r = 10$$ and $$r = 11$$ give the maximum and are equal:

$$c = \;^{21}C_{10} = \;^{21}C_{11}.$$

Using the recursive relation $$^{n}C_{r} = \;^{\,n-1}C_{r} + \;^{\,n-1}C_{r-1},$$

$$\begin{aligned} c &= ^{21}C_{10} \\ &= ^{20}C_{10} + \;^{20}C_{9} \\ &= 184756 + 167960 \\ &= 352716. \end{aligned}$$

We have therefore obtained

$$a = 92378,\qquad b = 184756,\qquad c = 352716.$$

Now we examine the ratios suggested by the four options. Start by dividing each number by the respective denominators in Option C:

$$ \begin{aligned} \dfrac{a}{11} &= \dfrac{92378}{11} = 8398,\\[4pt] \dfrac{b}{22} &= \dfrac{184756}{22} = 8398,\\[4pt] \dfrac{c}{42} &= \dfrac{352716}{42} = 8398. \end{aligned} $$

All three quotients are exactly equal to $$8398$$, so the three fractions are equal:

$$\frac{a}{11} = \frac{b}{22} = \frac{c}{42}.$$

No other option satisfies this equality; for example, under Option A we would have

$$\frac{c}{21}=16796 \neq 8398=\frac{a}{11}.$$

Hence, the correct answer is Option C.