Let $$0 \leq r \leq n$$. If $$^{n+1}C_{r+1} : ^{n}C_{r} : ^{n-1}C_{r-1} = 55 : 35 : 21$$, then $$2n + 5r$$ is equal to:

$$\frac{C(n+1,r+1)}{C(n,r)} = \frac{55}{35} = \frac{11}{7}$$. $$\frac{n+1}{r+1} = 11/7 \Rightarrow 7n+7 = 11r+11 \Rightarrow 7n = 11r+4$$.

$$\frac{C(n,r)}{C(n-1,r-1)} = \frac{35}{21} = \frac{5}{3}$$. $$\frac{n}{r} = 5/3 \Rightarrow n = 5r/3$$.

From second: $$r = 3n/5$$. Sub in first: $$7n = 11(3n/5)+4 = 33n/5+4$$. $$35n-33n = 20$$. $$2n = 20$$, $$n = 10$$. $$r = 6$$.

$$2n+5r = 20+30 = 50$$.

The correct answer is Option (1): 50.