If \({ }^{\mathrm{n}} \mathrm{C}_{4},{ }^{\mathrm{n}} \mathrm{C}_{5}\) and \({ }^{\mathrm{n}} \mathrm{C}_{6}\) are in A.P then the value of \(\mathrm{n}\) can be (a) 14 (b) 11 (c) 9 (d) 5

For any positive integers n and r, where r ≤ n, the formula for combinations is given by: \[ { }^nC_r = \frac{n!}{r!(n-r)!}. \]

Using the formula mentioned in step 1, write the expressions for \({ }^{\mathrm{n}} \mathrm{C}_{4}\), \({ }^{\mathrm{n}} \mathrm{C}_{5}\) and \({ }^{\mathrm{n}} \mathrm{C}_{6}\): \[ { }^nC_4 = \frac{n!}{4!(n-4)!} , \ { }^nC_5 = \frac{n!}{5!(n-5)!} , \ { }^nC_6 = \frac{n!}{6!(n-6)!}. \]

Since \({ }^nC_4\), \({ }^nC_5\), \({ }^nC_6\) are in A.P., we have: \[ { }^nC_5 - { }^nC_4 = { }^nC_6 - { }^nC_5. \] Now, substitute the expressions we wrote in step 2 and simplify the equation as follows: \[ \frac{n!}{5!(n-5)!} - \frac{n!}{4!(n-4)!} = \frac{n!}{6!(n-6)!} - \frac{n!}{5!(n-5)!}. \]

To simplify the equation from step 3, cancel out the n! term from all terms: \[ \frac{1}{5!(n-5)!} - \frac{1}{4!(n-4)!} = \frac{1}{6!(n-6)!} - \frac{1}{5!(n-5)!}. \] Now, we can cross multiply and simplify further: \[ \frac{(n-6)!}{5!(n-5)!(n-6)!} - \frac{(n-5)!}{4!(n-4)!(n-5)!} = \frac{(n-5)!}{6!(n - 5)!(n-6)!} - \frac{(n-4)!}{5!(n-5)!(n-4)!}. \] After canceling out the common terms, we get: \[ \frac{1}{5(n-5)} - \frac{1}{4(n-4)} = \frac{1}{6(n-6)} - \frac{1}{5(n-5)}. \] Further simplifying the equation: \[ \frac{n-4}{5(n-5)(n-4)} - \frac{n-5}{4(n-5)(n-4)} = \frac{n-5}{6(n-6)(n-5)} - \frac{n-6}{5(n-5)(n-6)}. \] Since all denominators are positive, we can cancel them out and simplify to: \[ (n-4)5 - 4(n-5) = 5(n-6) - 6(n-5). \] Expanding and solving for n: \[ 5n - 20 - 4n + 20 = 5n - 30 - 6n + 30. \] Rearranging terms gives: \[ n = 5. \] The value of n can be 5 (Option d).