Maths Solve the Following Question.(3 Marks)

$\begin{array}{ll}\therefore & r !=6 \\ \therefore & r=3\end{array}$

Substituting $r=3$ in (i), we get

$\begin{array}{ll} & \frac{n !}{(n-3) !}=720 \\ \therefore \quad & \frac{n(n-1)(n-2)(n-3) !}{(n-3) !}=720 \\ \therefore \quad & n(n-1)(n-2)=10 \times 9 \times 8 \\ \therefore \quad & n=10\end{array}$

$\therefore$ Required number of numbers formed $=\frac{6 !}{2 !}$

$=\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}$

= 360 A 6-digit number is to be formed using the same digits that are divisible by 4. For a number to be divisible by 4, the last two digits should be divisible by 4, i.e. 24, 52, 56, 64, 92 or 96. Case I: When the last two digits are 24, 52, 56 or 64. As the digit 9 repeats twice in the remaining four numbers, the number of arrangements =

$\frac{4 !}{2 !}=12$

∴ 6-digit numbers that are divisible by 4 so formed are 12 + 12 + 12 + 12 = 48. Case II: When the last two digits are 92 or 96. As each of the remaining four numbers are distinct, the number of arrangements = 4! = 24 ∴ 6-digit numbers that are divisible by 4 so formed are 24 + 24 = 48. ∴ Required number of numbers framed = 48 + 48 = 96

II. 2 books are together. Let us consider two books as one unit. This unit with the other 3 books can be arranged in${ }^4 P_4=4 !=24$ ways.

Also, two books can be arranged among themselves in ${ }^2 P _2=2$ ways.

∴ Required number of arrangements = 24 × 2 = 48

III. Say books are $B_1, B_2, B_3, B_4, B_5$ are to be arranged with $B_1, B_2$ never together.

$B_3, B_4, B_5$ can be arranged among themselves in ${ }^3 P_3=3 !=6$ ways.

$B_3, B_4, B_5$ create 4 gaps in which $B_1, B_2$ are arranged in ${ }^4 P_2=4 \times 3=12$ ways.

∴ Required number of arrangements = 6 × 12 = 72