A = ($$-$$$$\infty$$, 1) $$\cup$$ (3, $$\infty$$)

B = ($$-$$$$\infty$$, $$-$$2) $$\cup$$ (2, $$\infty$$)

C = ($$-$$$$\infty$$, 2] $$\cup$$ [6, $$\infty$$)

So, A $$\cap$$ B $$\cap$$ C = ($$-$$$$\infty$$, $$-$$2) $$\cup$$ [6, $$\infty$$)

z $$\cap$$ (A $$\cap$$ B $$\cap$$ C)' = {$$-$$2, $$-$$1, 0, $$-$$1, 2, 3, 4, 5}

Hence, no. of its subsets = 2⁸ = 256.

F(mn) = f(m) . f(n)

Put m = 1 f(n) = f(1) . f(n) $$\Rightarrow$$ f(1) = 1

Put m = n = 2

$$f(4) = f(2).f(2)\left\{ \matrix{ f(2) = 1 \Rightarrow f(4) = 1 \hfill \cr or \hfill \cr f(2) = 2 \Rightarrow f(4) = 4 \hfill \cr} \right.$$

Put m = 2, n = 3

$$f(6) = f(2).f(3)\left\{ \matrix{ when\,f(2) = 1 \hfill \cr f(3) = 1\,to\,7 \hfill \cr \hfill \cr f(2) = 2 \hfill \cr f(3) = 1\,or\,2\,or\,3 \hfill \cr} \right.$$

f(5), f(7) can take any value

Total = (1 $$\times$$ 1 $$\times$$ 7 $$\times$$ 1 $$\times$$ 7 $$\times$$ 1 $$\times$$ 7) + (1 $$\times$$ 1 $$\times$$ 3 $$\times$$ 1 $$\times$$ 7 $$\times$$ 1 $$\times$$ 7)

= 490

B $$-$$ C $$ \equiv $$ {7, 13, 19, ......, 97, .......}

Now, n² $$-$$ n $$\le$$ 100 $$\times$$ 100

$$\Rightarrow$$ n(n $$-$$ 1) $$\le$$ 100 $$\times$$ 100

$$\Rightarrow$$ A = {1, 2, ......., 100}.

So, A$$\cap$$(B $$-$$ C) = {7, 13, 19, ......., 97}

Hence, sum = $${{16} \over 2}(7 + 97) = 832$$

f(1) + f(2) = 3 $$-$$ f(3)

$$\Rightarrow$$ f(1) + f(2) = 3 + f(3) = 3

The only possibility is : 0 + 1 + 2 = 3

$$\Rightarrow$$ Elements 1, 2, 3 in the domain can be mapped with 0, 1, 2 only.

So number of bijective functions.

$$\left| \!{\underline {\, 3 \,}} \right. $$ $$\times$$ $$\left| \!{\underline {\, 5 \,}} \right. $$ = 720