Number of matches = $^{14}C_2 = \frac{14 \times 13}{2 \times 1} = 91$.

Total digits from 0 to 9 = 10, but digits at the first place = 9, i.e., number of digits = 9.
Required number = $9 \times P(9, 7) = 9 \times \frac{9!}{(9-7)!} = 9 \times \frac{9!}{2!}$.

$^{n}P_{4} = 12 \times ^{n}P_{2}$
$\frac{n!}{(n-4)!} = 12 \times \frac{n!}{(n-2)!}$
$\frac{1}{(n-4)!} = 12 \times \frac{1}{(n-2)(n-3)(n-4)!}$
$(n-2)(n-3) = 12$
$n^2 - 5n + 6 = 12$
$n^2 - 5n - 6 = 0$
$(n-6)(n+1) = 0 \Rightarrow n=6$.

If $^{n}C_{10} = ^{n}C_{15}$, then $n = 10 + 15 = 25$.
Therefore, as per question, $^{27}C_n = ^{27}C_{25} = ^{27}C_2 = \frac{27 \times 26}{2 \times 1} = 351$.

Let total participants = $n$.
If two players fall ill after playing four games each, then remaining participants = $(n-2)$ and total games = 53.
Therefore, as per question - $(n-2)C_2 + 8 = 53$
$\Rightarrow (n-2)C_2 = 45$
$\Rightarrow \frac{(n-2)(n-3)}{2} = 45$
$\Rightarrow (n-2)(n-3) = 90 = 10 \times 9$
$n-2 = 10 \Rightarrow n=12$.

Ways for m men to sit = $m!$
Places for n women = $m+1$
Ways = $m! \times ^{(m+1)}C_n \times n! = \frac{m! (m+1)!}{(m-n+1)!}$.

Total balls = 2+3+4=9. Number of combinations formed by picking 3 balls from 9 = $^9C_3 = 84$.
If black balls are left out, then total balls = (2+4)=6.
Now, the number of combinations formed by picking 3 balls from 6, in which there is not a single black ball, will be $^6C_3$.
$^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

Types of committee:
1 woman 2 men = $^5C_1 \times ^4C_2 = 5 \times 6 = 30$
2 women 1 man = $^5C_2 \times ^4C_1 = 10 \times 4 = 40$
3 women 0 men = $^5C_3 \times ^4C_0 = 10 \times 1 = 10$
Total ways = $30 + 40 + 10 = 80$.

Situations to form groups with at least one girl:
(1 girl, 3 boys) + (2 girls, 2 boys) + (3 girls, 1 boy) + (4 girls, 0 boys)
$= (^6C_1 \times ^4C_3) + (^6C_2 \times ^4C_2) + (^6C_3 \times ^4C_1) + (^6C_4 \times ^4C_0)$
$= (6 \times 4) + (15 \times 6) + (20 \times 4) + (15 \times 1)$
$= 24 + 90 + 80 + 15 = 209$