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: Apply Pascal's rule.
Pascal's identity says ${}^{n-1}C_3+{}^{n-1}C_4={}^{n}C_4$.
Step 2: Rewrite the inequality.
So the condition ${}^{n-1}C_3+{}^{n-1}C_4>{}^{n}C_3$ becomes ${}^{n}C_4>{}^{n}C_3$.
Step 3: Form the ratio.
$\dfrac{{}^{n}C_4}{{}^{n}C_3}=\dfrac{n-3}{4}$.
Step 4: Set the ratio above one.
We need $\dfrac{n-3}{4}>1$, that is $n-3>4$.
Step 5: Solve for $n$.
This gives $n>7$.
Step 6: Pick the least valid value.
The smallest integer greater than $7$ is $8$, which is option (3).
\[ \boxed{8} \]