If displaystyle ∑ r = 025 ( 50Cr) . ( 50 - rC25 - r) =K( 50C25), then K is equal to

Question

If displaystyle ∑ r = 025 ( 50Cr) . ( 50 - rC25 - r) =K( 50C25), then K is equal to (A) 225 (B) 225-1 (C) 224 (D) (25)2

Tardigrade Admin · Accepted Answer

LHS = displaystyle ∑ r = 025 50 Cr 50 - rC25 - r = displaystyle ∑ r = 025 (50 !/r ! (50 - r) !) ((50 - r) !/(25 - r) ! (25) !) = displaystyle ∑ r = 025 (50 ! 25 !/r ! (25 - r) ! (25) ! (25) !) = 50C25 displaystyle ∑ r = 025 25 Cr = 50C25 × 225 ⇒ 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

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=225

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

$2^{25}$

$2^{25} - 1$

$2^{24}$

$(25)^{2}$