Let S = {1, 2, 3, .........., 9}. For k = 1, 2, .........., 5, let N_k be the number of subsets of S, each containing five elements out of which exactly k are odd. Then N₁ + N₂ + N₃ + N₄ + N₅ =

Let O be the origin and let PQR be an arbitrary triangle. The point S is such that

$$\overrightarrow{OP}$$ . $$\overrightarrow{OQ}$$ + $$\overrightarrow{OR}$$ . $$\overrightarrow{OS}$$ = $$\overrightarrow{OR}$$ . $$\overrightarrow{OP}$$ + $$\overrightarrow{OQ}$$ . $$\overrightarrow{OS}$$ = $$\overrightarrow{OQ}$$ . $$\overrightarrow{OR}$$ + $$\overrightarrow{OP}$$ . $$\overrightarrow{OS}$$

Then the triangle PQR has S as its

The equation of the plane passing through the point (1, 1, 1) and perpendicular to the planes 2x + y $$-$$ 2z = 5 and 3x $$-$$ 6y $$-$$ 2z = 7 is

f : R $$ \to $$ R is a differentiable function such that f'(x) > 2f(x) for all x$$ \in $$R, and f(0) = 1 then