Question

The least value of \(n\) so that \[ {}^{\,n-1}C_3+{}^{\,n-1}C_4\gt {}^{\,n}C_3 \] is

• A. \(11\)
• B. \(9\)
• C. \(8\)
• D. \(7\)

Hint:

Use Pascal's identity: \[ {}^{n-1}C_r+{}^{n-1}C_{r-1}={}^{n}C_r \] to simplify sums of combinations quickly.

Answer 1

The Correct Option is C

Step 1: Use Pascal's identity.
We know that \[ {}^{n}C_r={}^{n-1}C_r+{}^{n-1}C_{r-1} \] For \(r=4\), \[ {}^{n}C_4={}^{n-1}C_4+{}^{n-1}C_3 \] Thus, \[ {}^{\,n-1}C_3+{}^{\,n-1}C_4={}^{n}C_4 \] So the given inequality becomes \[ {}^{n}C_4\gt {}^{n}C_3 \]

Step 2: Write the combination formulas.
\[ {}^{n}C_4=\frac{n!}{4!(n-4)!} \] and \[ {}^{n}C_3=\frac{n!}{3!(n-3)!} \] Now compare: \[ {}^{n}C_4\gt {}^{n}C_3 \]

Step 3: Use the ratio method.
\[ \frac{{}^{n}C_4}{{}^{n}C_3} = \frac{\frac{n!}{4!(n-4)!}}{\frac{n!}{3!(n-3)!}} \] \[ = \frac{3!(n-3)!}{4!(n-4)!} \] \[ = \frac{n-3}{4} \] For \[ {}^{n}C_4\gt {}^{n}C_3, \] we need \[ \frac{n-3}{4}\gt 1 \]

Step 4: Solve the inequality.
\[ n-3\gt 4 \] \[ n\gt 7 \] Since \(n\) is a natural number, the least possible value is \[ n=8 \]

Step 5: Final conclusion.
Hence, \[ \boxed{8} \] which corresponds to option (3).