Permutations & Combinations
Grade 11
Question:
<p>If <span class="math-tex">\(^n C_4, \ ^nC_5 \ and \ ^n C_6\)</span> are in AP, then n can be</p>
<p style="display:inline">11</p>
<p style="display:inline">12</p>
<p style="display:inline">9</p>
<p style="display:inline">14</p>
Step-by-Step Solution
Key Concept: Apply the arithmetic progression property 2b = a + c to the binomial coefficients and simplify the resulting equation by cancelling common factorial terms.
<p>If <sup>n</sup>C<sub>4</sub>, <sup>n</sup>C<sub>5 </sub>and <sup>n</sup>C<sub>6</sub> are in AP, then<br />
<span class="math-tex">\(2 \cdot^{n} C_{5}=^{n} C_{4}+^{n} C_{6}\)</span><br />
[If a, b, c are in AP , then 2b = a + c]<br />
<span class="math-tex">\(\Rightarrow 2 \frac{n !}{5 !(n-5) !}=\frac{n !}{4 !(n-4) !}+\frac{n !}{6 !(n-6) !}\)</span><br />
<span class="math-tex">\(\left[\because^{n} C_{r}=\frac{n !}{r !(n-r) !}\right]\)</span><br />
<span class="math-tex">\(\Rightarrow \frac{2}{5 \cdot 4 !(n-5)(n-6) !}\)</span><br />
<span class="math-tex">\(=\frac{1}{4 !(n-4)(n-5)(n-6) !}+\frac{1}{6 \cdot 5 \cdot 4 !(n-6) !}\)</span><br />
<span class="math-tex">\(\Rightarrow \quad \frac{2}{5(n-5)}=\frac{1}{(n-4)(n-5)}+\frac{1}{30}\)</span><br />
<span class="math-tex">\(\Rightarrow\)</span> 12 (n - 4) = 30 + n<sup>2</sup> - 9n + 20<br />
<span class="math-tex">\(\Rightarrow\)</span> n<sup>2</sup> - 21n + 98 = 0<br />
<span class="math-tex">\(\Rightarrow\)</span> n<sup>2</sup> - 14n - 7n + 98 = 0<br />
<span class="math-tex">\(\Rightarrow\)</span> n(n - 14) - 7(n - 14) = 0<br />
<span class="math-tex">\(\Rightarrow\)</span> (n - 7) (n - 14) = 0<br />
<span class="math-tex">\(\Rightarrow\)</span> n = 7 or 14</p>
Correct Answer: D