Suppose  $$A$$  and  $$B$$  are the coefficients of $$30^{\text{th}}$$  and $$12^{\text{th}}$$  terms respectively in the binomial expansion of  $$(1+x)^{2n-1}.$$ If $$2A=5B,$$  then  $$n$$  is equal to: 

Identify coefficients

In $$(1+x)^{2n-1}$$:

• $$A = T_{30} \text{ coeff} = \binom{2n-1}{29}$$
• $$B = T_{12} \text{ coeff} = \binom{2n-1}{11}$$

Step 2: Use the ratio 2A = 5B

$$2 \cdot \frac{(2n-1)!}{29!(2n-30)!} = 5 \cdot \frac{(2n-1)!}{11!(2n-12)!}$$

Cancel $$(2n-1)!$$ and rearrange:

$$\frac{2}{5} = \frac{29!}{11!} \cdot \frac{(2n-30)!}{(2n-12)!}$$

$$\frac{2}{5} = \frac{29 \cdot 28 \cdot ... \cdot 12}{(2n-12) \cdot (2n-13) \cdot ... \cdot (2n-29)}$$

Alternatively, using the property $$\binom{n}{r} = \binom{n}{n-r}$$, if we test $$n=21$$:

$$2 \binom{41}{29} = 5 \binom{41}{11}$$

Since $$\binom{41}{29} = \binom{41}{41-29} = \binom{41}{12}$$:

$$2 \binom{41}{12} = 5 \binom{41}{11} \implies 2 \frac{41!}{12!29!} = 5 \frac{41!}{11!30!}$$

$$2/12 = 5/30 \implies 1/6 = 1/6$$. (Matches!)

Correct Option: C (21)