Animated Solution for Mathematics - Binomial Theorem: If $a,b$ and $c$ are the greatest values of ${}^{19}{C}_p{,}^{20}{C}_q{,}^{21}{C}_r$ respectively, then :

The binomial coefficients ${}^n{C}_r$ follow a specific pattern of growth and decay. For any row $n$, the coefficients increase until they reach the middle and then decrease symmetrically.

If $n$ is even, there is a single maximum value at $r=\frac{n}{2}$. If $n$ is odd, the symmetry creates a plateau where the two middle terms, $r=\frac{n-1}{2}$ and $r=\frac{n+1}{2}$, share the maximum value.

For $a=\max {(}^{19}{C}_p)$, since $19$ is odd, the peak occurs at $p=9$ or $10$. We choose $a{=}^{19}{C}_9$.

For $b=\max {(}^{20}{C}_q)$, since $20$ is even, the peak is singular at $q=10$. Thus, $b{=}^{20}{C}_{10}$.

For $c=\max {(}^{21}{C}_r)$, since $21$ is odd, the peak occurs at $r=10$ or $11$. We choose $c{=}^{21}{C}_{11}$.

To relate these values without calculating large factorials, we utilize the fundamental recurrence relation:

Applying this to $b{=}^{20}{C}_{10}$, we can express it in terms of $a$:

This simplifies to the ratio:

Next, we apply the same logic to $c{=}^{21}{C}_{11}$ in terms of $b$:

This rearranges to the ratio:

We now have two ratios: $\frac{a}{1}=\frac{b}{2}$ and $\frac{b}{11}=\frac{c}{21}$. To unite them, we find a common denominator for $b$.

Multiplying the first ratio by $11$ gives:

Multiplying the second ratio by $2$ gives:

By chaining these together, we arrive at the final, harmonious relationship:

This result demonstrates that the maximum values of these binomial rows are proportional to the values $11,22,$ and $42$.