- A$\frac{121}{10^{4}}$
- B$\frac{150}{10^{4}}$
- ✓$\frac{135}{10^{4}}$
- D$\frac{134}{10^{4}}$
Now, number of 5 -digit numbers containing both digits $=2^{5}-2$
Second Case: Choose one non-zero \& one zero as digit ${ }^{9} C _{1}$
Number of 5 -digit numbers containg one non zero and one zero both $=\left(2^{4}-1\right)$ Required prob.
$=\frac{\left({ }^{9} C _{2} \times\left(2^{5}-2\right)+{ }^{9} C _{1} \times\left(2^{4}-1\right)\right)}{9 \times 10^{4}}$
$=\frac{36 \times(32-2)+9 \times(16-1)}{9 \times 10^{4}}$
$=\frac{4 \times 30+15}{10^{4}}=\frac{135}{10^{4}}$
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Statement $-1$ : Equation of other roots is $x^2 + cx + ab = 0$
Statement $-2$ : $a + b + c = 0$
$x^2 + y^2 = 1 ; x^2 + y^2 + 2x - 3 = 0$ and $x^2 + y^2 + 2y - 3 = 0$ is