^47C4 + ∑ {r=1 to 5} ^(52-r)C3 is equal to

Question

⁴⁷C₄  + \(\sum_{r=1}^{5}\) \(^{52-r}C_3\)  is equal to 

(a) ⁴⁷C₆ 

(b) ⁵²C₅ 

(c) ⁵²C₄ 

(d) None of these

Answer 1

Answer :(c) ⁵²C₄

\(^{47}C_4 + \sum_{r=1}^5 \)\(^{52-r}C_3\) 

= ⁵¹C₃ + ⁵⁰C₃ + ⁴⁹C₃ + ⁴⁸C₃ + ⁴⁷C₃ + ⁴⁷C₄ 

= ⁵¹C₃ + ⁵⁰C₃ + ⁴⁹C₃ + ⁴⁸C₃ + ⁴⁸C₄  (\(\because\) ⁿC_r + ⁿC_r+1 = ⁿ⁺¹C_r+1) 

= ⁵¹C₃ + ⁵⁰C₃ + ⁴⁹C₃ + ⁴⁹C₄ 

= ⁵¹C₃ + ⁵⁰C₃ + ⁵⁰C₄ 

= ⁵¹C₃ + ⁵¹C₄ 

= ⁵²C₄