The correct option is A (3010)
To find
30C030C10−30C301C11+30C302C12−...+30C20 30C30
We know that
(1+x)30=30C0+30C1x+30C2x2+.....+30C20x20+...30C30x30 .....(1)(x−1)30=30C0x30−30C1x29+....+30C10x20−30C11x19+30C12x18+...30C30x0 .....(2)
Multiplying equation (1) and (2),we get
(x2−1)30=()×()
Equating the coefficients of x20 on both sides, we get
30C10=30C300C10−30C301C11+30C302C12−...+30C20 30C30
∴ Req.value is 30C10