If displaystyle∑r=025 ( 50 Cr)( 50-r C25-r) =K( 50 C25), then K is equal to

Question

If displaystyle∑r=025 ( 50 Cr)( 50-r C25-r) =K( 50 C25), then K is equal to (A) 225 (B) 225-1 (C) 224 (D) (25)2

Tardigrade Admin · Accepted Answer

displaystyle∑r=025 ( 50 Cr)( 50-r C25-r) =K( 50 C25) LHS = displaystyle∑r=025 50 Cr 50-r C25-r = displaystyle∑r=025[ (50 !/r !(50-r) !) ((50-r) !/(25-r) !(25) !) ] = displaystyle∑r=025(50 ! 25 !/r !(25-r) !(25) !(25) !) = 50 C25 displaystyle∑r=025 25 Cr = 50 C25 × 225 =K( 50 C25) ∴ K=225

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

$2^{25}$

$2^{25} - 1$

$2^{24}$

$(25)^{2}$

Q. If r=0∑25​{(50Cr​)(50−rC25−r​)}=K(50C25​), then K is equal to

225

225−1

224

(25)2

r=0∑25​{(50Cr​)(50−rC25−r​)}=K(50C25​) LHS =r=0∑25​50Cr​50−rC25−r​ =r=0∑25​[{r!(50−r)!50!​}{(25−r)!(25)!(50−r)!​}] =r=0∑25​r!(25−r)!(25)!(25)!50!25!​ =50C25​r=0∑25​25Cr​ =50C25​×225 =K(50C25​) ∴K=225

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

$2^{25}$

$2^{25} - 1$

$2^{24}$

$(25)^{2}$