(17280)

Number of ways that all boys sit together $=5!\times 5!$

Number of ways no two boys sit together $=4!\times 5!$

$\therefore$   Required number of ways $=5!\times 5!+4!\times 5!=17280$

(125)

Number of 3-digits = 999 – 99 = 900

Number of 3-digit numbers divisible by 2 & 3 i.e., by 6,

        $\frac{900}{6}=150$

Number of 3-digit numbers divisible by 4 & 9 i.e., by 36,

         $\frac{900}{36}=25$

$\therefore$  Number of 3-digit numbers divisible by 2 & 3 but not 4 & 9 = 150 – 25 = 125.

(392)

Given : $f:\{1,2,...,100\}\to \{0,1\}$

Number of ways to connect {1, 2, ..., 98} to 1 = 98

Number 99 can connect either 0 or 1 $\Rightarrow$ 2 ways

Similarly, 100 can connect either 0 or 1 $\Rightarrow$ 2 ways

$\therefore$   Total number of functions for the given condition that assign 1 to exactly one of positive integers $\le$ 98 is given by $98\times 2\times 2$ = 392.

(64)

Let xyz be any number between 212 and 999

Let $x=2\to y+z=13$, then

(y, z) : (4, 9), (5, 8), (6, 7), (7, 6), (8, 5), (9, 4), i.e., 6 in number.

Let $x=3\to y+z=12$, then

(y, z) : (3, 9), (4, 8), (5, 7), (6, 6), (7, 5), (8, 4), (9, 3) i.e., 7 in number.

Let $x=4\to y+z=11$, then

(y, z) : (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4),(8, 3), (9, 2) i.e., 8 in number.

Let $x=5\to y+z=10$, then

(y, z) : (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3). (8, 2), (9, 1) i.e., 9 in number.

Let $x=6\to y+z=9$, then

(y, z) : (0, 9), (1, 8), (2, 7), (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1), (9, 0) i.e., 10 in number.

Let $x=7\to y+z=8$, then

(y, z) : (0, 8), (1, 7), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (7, 1), (8, 0) i.e., 9 in number.

Let $x=8\to y+z=7$, then

(y, z) : (0, 7), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6,1), (7, 0) i.e., 8 in number.

Let $x=9\to y+z=6$, then

(y, z) : (0, 6), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 0) i.e., 7 in number.

Total = 6 + 7 + 8 + 9 + 10 + 9 + 8 + 7 = 64.