Question
Mathematics Question on Combinations
The value of (030)(1030)−(130)(1130)+(230)(1230).....+(2030)(3030) is where (rn)=nCr
A
(1030)
B
(1530)
C
(3060)
D
(1031)
Answer
(1030)
Explanation
Solution
To find 30C030C10−30C130C11+30C230C12−....+30C2030C30 We know that (1+x)30=30C0+30C1x+30C2x2 +....+30C20x20+....30C30x30....(1) (x−1)30=30C0x30−30C1x29+....+30C10x20 −30C11x19+30C12x18+....30C30x0....(2) Multiplying eqn(1) and (2) and equating the coefficients of x20 on both sides, we get 30C10=30C030C10−30C130C11+30C230C12−....+30C2030C30 ∴ Re value is 30C10