Permutations and Combinations 2 Question 6
6. If $\sum _{r=0}^{25}{{ }^{50} C _r \cdot{ }^{50-r} C _{25-r} }=K\left({ }^{50} C _{25}\right)$,
then, $K$ is equal to
(a) $2^{24}$
(b) $2^{25}-1$
(c) $2^{25}$
(d) $(25)^{2}$
(2019 Main, 10 Jan II)
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Answer:
Correct Answer: 6. (c)
Solution:
- Given, $\sum _{r=0}^{25}{{ }^{50} C _r \cdot{ }^{50-r} C _{25-r} }=K{ }^{50} C _{25}$
$\Rightarrow \sum _{r=0}^{25} \frac{50 !}{r !(50-r) !} \times \frac{(50-r) !}{(25-r) ! 25 !}=K{ }^{50} C _{25}$ $\Rightarrow \quad \sum _{r=0}^{25} \frac{50 !}{25 ! 25 !} \times \frac{25 !}{r !(25-r) !}=K{ }^{50} C _{25}$
[on multiplying 25 ! in numerator and denominator.]
$ \begin{array}{ccc} \Rightarrow & { }^{50} C _{25} \sum _{r=0}^{25}{ }^{25} C _r=K & { }^{50} C _{25} \\ \Rightarrow & \because{ }^{50} C _{25}=\frac{50 !}{25 ! 25 !} \\ & {\left[\because{ } _{r=0}^{25}{ }^{25} C _r=2^{25}+{ }^{n} C _1+{ }^{n} C _2+\ldots .+{ }^{n} C _n=2^{n}\right]} \\ \Rightarrow & K=2^{25} \end{array} $
