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:

  1. 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} $



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