SATHEE: Permutations and Combinations 2 Question 6

Permutations and Combinations 2 Question 6

6. If $\sum_{r = 0}^{25}^{50} C_{r} \cdot^{50 - r} C_{25 - r} = K (^{50} C_{25})$ ,

then, $K$ is equal to

(a) $2^{24}$

(b) $2^{25} - 1$

(d) $(25)^{2}$

(2019 Main, 10 Jan II)

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Answer:

Correct Answer: 6. (c)

Solution:

Given, $\sum_{r = 0}^{25}^{50} C_{r} \cdot^{50 - r} C_{25 - r} = K^{50} C_{25}$

$\Rightarrow \sum_{r = 0}^{25} \frac{50!}{r! (50 - r)!} \times \frac{(50 - r)!}{(25 - r)! 25!} = K^{50} C_{25}$ $\Rightarrow \sum_{r = 0}^{25} \frac{50!}{25! 25!} \times \frac{25!}{r! (25 - r)!} = K^{50} C_{25}$

[on multiplying 25 ! in numerator and denominator.]

$\begin{array}{ccc} \Rightarrow & ^{50} C_{25} \sum_{r = 0}^{25}^{25} C_{r} = K & ^{50} C_{25} \\ \Rightarrow & ∵^{50} C_{25} = \frac{50!}{25! 25!} \\ [∵_{r = 0}^{25}^{25} C_{r} = 2^{25} +^{n} C_{1} +^{n} C_{2} + \dots . +^{n} C_{n} = 2^{n}] \\ \Rightarrow & K = 2^{25} \end{array}$

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Permutations and Combinations 2 Question 6

6. If ∑r=02550Cr⋅50−rC25−r=K(50C25),

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6. If $\sum_{r = 0}^{25}^{50} C_{r} \cdot^{50 - r} C_{25 - r} = K (^{50} C_{25})$ ,