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EEJ MAIN Mathematics QUESTION #6938
Question 1

If $\displaystyle\sum_{r=0}^{25}\left\{{}^{50}C_r \cdot {}^{50-r}C_{25-r}\right\} = K\binom{50}{25}$, find the value of $K$.

  • $(25)^2$
  • $2^{25}-1$
  • $2^{24}$
  • $2^{25}$✔️
Correct Answer Explanation

Note: ${}^{50}C_r \cdot {}^{50-r}C_{25-r} = \dfrac{50!}{r!(50-r)!} \cdot \dfrac{(50-r)!}{(25-r)!(25)!} = \dfrac{50!}{r!(25-r)!25!} = {}^{50}C_{25}\cdot{}^{25}C_r$

So the sum $= {}^{50}C_{25}\displaystyle\sum_{r=0}^{25}{}^{25}C_r = {}^{50}C_{25}\cdot 2^{25}$

Therefore $K = \mathbf{2^{25}}$.