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Answer: $\mathbf C$

Explanation:

$\binom{n}{r} = 84 \implies \frac{n!}{(n-r)!r!} = 84$

$\binom{n}{r-1} = 84 \implies \frac{n!}{(n-r+1)!(r-1)!} = 84$

$\binom{n}{r+1} = 84 \implies \frac{n!}{(n-r-1)!(r+1)!} = 126$

$\frac{\binom{n}{r}}{\binom{n}{r-1}}=\frac{84}{36} = \frac{7}{3} \implies \frac{n!}{(n-r)!r!} .\frac{(n-r+1)!(r-1)!}{n!} = \frac{n-r+1}{r} = \frac{7}{3}$

On simplifying this, we get:

$3n + 3 = 10r \cdots(1)$

Similarly,

$\frac{\binom{n}{(r+1)}}{\binom{n}{r}}=\frac{126}{84} = \frac{3}{2} \implies \frac{n-r}{r+1} = \frac{3}{2}$

Again, on simplifying this, we get:

$2n-3 = 5r \cdots(2)$

On solving equation $(1)$ and $(2)$, we get:

$r= 3$

$\therefore \mathbf C$ is the correct option.

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