Example 9. If
$$ \sum_{r=0}^{25} \binom{50}{r} \cdot \binom{50-r}{25-r} = K \binom{50}{25}, $$
then, K is equal to
(a) $$2^{24}$$
(b) $$2^{25} – 1$$
(c) $$2^{25}$$
(d) $$(25)^2$$
(JEE Main 2019)
Solution: (c) Given,
$$ \sum_{r=0}^{25} \binom{50}{r} \cdot \binom{50-r}{25-r} = K \binom{50}{25} $$
Expanding binomial coefficients:
$$ \sum_{r=0}^{25} \frac{50!}{r!(50-r)!} \times \frac{(50-r)!}{(25-r)! 25!} = K \binom{50}{25} $$
Simplifying,
$$ \sum_{r=0}^{25} \frac{50!}{r!(25-r)! 25!} = K \binom{50}{25} $$
Multiplying and dividing by 25!:
$$ \binom{50}{25} \sum_{r=0}^{25} \binom{25}{r} = K \binom{50}{25} $$
Since
$$ \sum_{r=0}^{25} \binom{25}{r} = 2^{25} $$
Thus,
$$ K = 2^{25} $$
Final Answer:
$$ K = 2^{25} $$
Inline Equation Example: Albert Einstein’s equation is $E = mc^2$.
Block Equation Example:
$$
E = mc^2
$$
Inline Equation Example: Albert Einstein’s equation is (E = mc^2).
Block Equation Example:
$$
E = mc^2
$$
Inline Equation Example: Albert Einstein’s equation is ( \displaystyle E = mc^2 ).
Block Equation Example:
$$
E = mc^2
$$
