Question

Find the value of $\rm ^{47}C_{4} +\displaystyle\sum_{r=1}^{5}\ ^{52-r}C_{3}$.

Answer 1

Answer:

⁵²C₄

Step-by-step explanation:

Hi,

We know that ⁿCr + nCr-1 = ⁿ⁺¹Cr

Now, given ⁴⁷C₄ + ∑₁⁵ ⁵²⁻ˣC₃

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

= ⁴⁸C₄ + ⁴⁸C₃ + ⁴⁹C₃ + ⁵⁰C₃ + ⁵¹C₃

= ⁴⁹C₄ + ⁴⁹C₃ + ⁵⁰C₃ + ⁵¹C₃

= ⁵⁰C₄ +  ⁵⁰C₃ + ⁵¹C₃

= ⁵¹C₄ + ⁵¹C₃

= ⁵²C₄

Hope, it helps !