JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023The value of ∑r=02222Cr23Cr is
Choose the correct answer:
- A.
44C23
- B.
45C23
(Correct Answer) - C.
44C22
45C23
Explanation
Let S=∑r=022(22Cr)(23Cr)
Consider, (1+x)22=22C0x0+22C1x1+⋯+22C22x22…(1) Again, (x+1)23=23C0x23+23C1x22+⋯+23C23x0…(2)
(1)×(2) gives
(1+x)45=(22C0x0+22C1x1+⋯+22C22x22)×(23C0x23+23C1x22+⋯+23C23x0)…(3)
Again, S=∑r=022(22C22−r)(23Cr),
Using nCr=nCn−r ⇒S=(22C22)(23C0)+(22C21)(23C1)+⋯+(22C0)(23C22)
From (3), comparing coefficient of x23 on both sides
or S=45C23
Explanation
Let S=∑r=022(22Cr)(23Cr)
Consider, (1+x)22=22C0x0+22C1x1+⋯+22C22x22…(1) Again, (x+1)23=23C0x23+23C1x22+⋯+23C23x0…(2)
(1)×(2) gives
(1+x)45=(22C0x0+22C1x1+⋯+22C22x22)×(23C0x23+23C1x22+⋯+23C23x0)…(3)
Again, S=∑r=022(22C22−r)(23Cr),
Using nCr=nCn−r ⇒S=(22C22)(23C0)+(22C21)(23C1)+⋯+(22C0)(23C22)
From (3), comparing coefficient of x23 on both sides
or S=45C23

