$$\begin{aligned} \text { Required number of ways } & ={ }^5 \mathrm{C}_2 \times{ }^7 \mathrm{C}_1+{ }^5 \mathrm{C}_3 \quad \text{[since, at least two red]}\\ & =10 \times 7+10 \\ & =70+10=80 \end{aligned}$$

Among the digits $0,1,2,3,4,5,6,7,8,9$, clearly $1,3,5,7$ and 9 are odd.

$\therefore$ Number of six-digit numbers $=5 \times 5 \times 5 \times 5 \times 5 \times 5=5^6$

Let the number of team participating in championship be $n$.

Since, it is given that every two teams played one match with each other.

$\therefore$ Total match played $={ }^n C_2$

According to the question,

$$\begin{array}{r} & { }^n C_2 =153 \\ \Rightarrow & \frac{n(n-1)}{2} =153 \\ \Rightarrow & n^2-n =306 \\ \Rightarrow & n^2-n-306 =0 \\ \Rightarrow & (n-18)(n+17) =0 \\ \Rightarrow & n =18,-17 \quad \text{ [inadmissible]}\\ \therefore & n =18 \end{array}$$

The arrangement can be understand with the help of following figure.

Thus, '+' sign can be arranged in 1 way because all are identical. and 4 negative signs can be arranged at 7 places in ${ }^7 C_4$ ways.

$$\begin{aligned} \therefore \quad \text { total number of ways } & ={ }^7 C_4 \times 1 \\ & =\frac{7!}{4!3!}=\frac{7 \times 6 \times 5 \times 4!}{3!\times 4!} \\ & =\frac{7 \times 6 \times 5}{3 \times 2 \times 1}=35 \text { ways } \end{aligned}$$

Since, there are 2 white, 3 black and 4 red balls. It is given that atleast one black ball is to be included in the draw.

$$\begin{aligned} \therefore \text { Required number of ways } & ={ }^3 \mathrm{C}_1 \times{ }^6 \mathrm{C}_2+{ }^3 \mathrm{C}_2 \times{ }^6 \mathrm{C}_1+{ }^3 \mathrm{C}_3 \\ & =3 \times 15+3 \times 6+1 \\ & =45+18+1=64 \end{aligned}$$