- r -025 C 50 C r -50- r C 25- r - K 50 C 25 then K is equal to -B- 224C- 225-1D- 225

The correct option is D 2^25∑_r=0 ^25 ^50C_r ^50 rC_25 r =∑_r=0 ^2550!r!(50 r)!×(50 r)!25!(25 r)! =∑_r=0 ^2550! 25!r! (25 r)! 25! 25! = ^50C_25∑_r=0 ^25 ^25C_r ...

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Byju's Answer

Standard IX

Mathematics

Quotient Law

∑ r =025 C 50...

Question

If $25 \sum r = 0 {^{50} C_{r} \cdot^{50 - r} C_{25 - r}} = K (^{50} C_{25})$ then $K$ is equal to :

25^{2}

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2^{24}

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2^{25} - 1

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2^{25}

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Solution

The correct option is D $2^{25}$
$25 \sum r = 0^{50} C_{r}^{50 - r} C_{25 - r}$
$= 25 \sum r = 0 \frac{50!}{r! (50 - r)!} \times \frac{(50 - r)!}{25! (25 - r)!}$
$= 25 \sum r = 0 \frac{50! 25!}{r! (25 - r)! 25! 25!}$
$=^{50} C_{25} 25 \sum r = 0^{25} C_{r}$
$=^{50} C_{25} \times 2^{25}$
Hence, $K = 2^{25}$

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Similar questions

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\sum_{r = 0}^{25} {^{50} C_{r} \cdot^{50 - r} C_{25 - r}} = K (^{50} C_{25})

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, then the remainder is

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If 25∑r=0{ 50Cr⋅ 50−rC25−r}=K( 50C25) then K is equal to :

The correct option is D 22525∑r=0 50Cr 50−rC25−r =25∑r=050!r!(50−r)!×(50−r)!25!(25−r)! =25∑r=050! 25!r! (25−r)! 25! 25! = 50C2525∑r=0 25Cr = 50C25×225 Hence, K=225

If $25 \sum r = 0 {^{50} C_{r} \cdot^{50 - r} C_{25 - r}} = K (^{50} C_{25})$ then $K$ is equal to :