If $$^nC_4$$, $$^nC_5$$ and $$^nC_6$$ are in A.P., then n can be

We have been told that the three binomial coefficients $$^nC_4,\; ^nC_5,\; ^nC_6$$ form an arithmetic progression. In an arithmetic progression the middle term is the average of its neighbours, so the defining condition is

$$2\,^nC_5 \;=\; ^nC_4 + ^nC_6.$$

First recall the factorial form of a binomial coefficient:

$$^nC_r \;=\; \dfrac{n!}{r!\,(n-r)!}.$$

There is also a very convenient ratio formula derived straight from the factorial definition:

$$^nC_{\,r+1} \;=\; \dfrac{n-r}{r+1}\; ^nC_r.$$ This will allow us to rewrite every term in the above equation in terms of the single quantity $$^nC_5$$.

To express $$^nC_4$$ in terms of $$^nC_5$$ we set $$r=4$$ in the ratio formula and obtain

$$^nC_5 \;=\; \dfrac{n-4}{5}\; ^nC_4 \;\;\Longrightarrow\;\; ^nC_4 \;=\; ^nC_5 \;\dfrac{5}{\,n-4}.$$

Next, to express $$^nC_6$$ in terms of $$^nC_5$$ we set $$r=5$$ in the same ratio formula and get

$$^nC_6 \;=\; \dfrac{n-5}{6}\; ^nC_5.$$

Now substitute these two expressions into the arithmetic-progression condition:

$$2\,^nC_5 \;=\; ^nC_5\,\dfrac{5}{n-4} \;+\; ^nC_5\,\dfrac{n-5}{6}.$$

Because $$^nC_5$$ is a common, non-zero factor, we may divide both sides by it:

$$2 \;=\; \dfrac{5}{n-4} \;+\; \dfrac{n-5}{6}.$$

To clear the denominators multiply every term by $$6\,(n-4):$$

$$2 \times 6\,(n-4) \;=\; 5 \times 6 \;+\; (n-5)\,(n-4).$$

Simplifying the left-hand side:

$$12\,(n-4).$$

Simplifying the right-hand side step by step gives

$$30 + (n-5)(n-4).$$

Expanding the product $$(n-5)(n-4)$$:

$$n^2 - 4n - 5n + 20 \;=\; n^2 - 9n + 20.$$

So the equation becomes

$$12(n-4) \;=\; 30 + n^2 - 9n + 20.$$

Combine the constants on the right:

$$12(n-4) \;=\; n^2 - 9n + 50.$$

Expand the left-hand side:

$$12n - 48 \;=\; n^2 - 9n + 50.$$

Now bring every term to the right to obtain a standard quadratic equation:

$$0 \;=\; n^2 - 9n + 50 - 12n + 48,$$

which simplifies to

$$n^2 - 21n + 98 \;=\; 0.$$

We solve this quadratic by the discriminant method. The discriminant is

$$\Delta \;=\; (-21)^2 - 4 \times 1 \times 98 \;=\; 441 - 392 \;=\; 49.$$

Since $$\sqrt{\Delta}=7,$$ the roots are

$$n \;=\; \dfrac{21 \pm 7}{2}.$$

This gives the two numerical values

$$n_1 = \dfrac{21 + 7}{2} = \dfrac{28}{2} = 14,$$ $$n_2 = \dfrac{21 - 7}{2} = \dfrac{14}{2} = 7.$$

Both numbers are indeed > 6, so both satisfy the algebraic condition. However, among the options offered (9, 14, 12, 11) only $$14$$ appears. Consequently the permissible value of $$n$$ from the given list is $$14$$.

Hence, the correct answer is Option B.