The number of five digit numbers divisible by 5 that can be formed using the numbers 0, 1, 2, 3, 4, 5 without repetition is

• 240

• 216

• 120

• 96

${}_{15}\mathrm{P}_8 = \mathrm{A} + 8 . {}_{14}\mathrm{P}_7 \Rightarrow \mathrm{A} =$

• ${}_{14}\mathrm{P}_6$

• ${}_{14}\mathrm{P}_8$

• ${}_{15}\mathrm{P}_7$

• ${}_{16}\mathrm{P}_9$

$\mathrm{If} {}^{\left(\mathrm{n} - 1\right)}\mathrm{C}_1 + {}^{\left(\mathrm{n} - 1\right)}\mathrm{C}_4 > {}^{\mathrm{n}}\mathrm{C}_3, \mathrm{then} \mathrm{the} \mathrm{minimum} \mathrm{value} \mathrm{of} \mathrm{n} \mathrm{is}$

• 5

• 6

• 7

• 8

$\mathrm{If} \mathrm{f}\left(\mathrm{x}\right) = {\left({\mathrm{x}}^2 - 1\right)}^7, \mathrm{then} {\mathrm{f}}^{\left(14\right)}\left(\mathrm{x}\right) = ?$

• 0

• 2!

• 7!

• 14!

$\mathrm{If} { }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r} - 1} = 330, {}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}} = 462, \mathrm{and}{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r} + 1} = 462, \mathrm{then} \mathrm{r} = ?$

• 3

• 4

• 5

• 6

10 men and 6 women are to be seated in a row so that no two women sit together. The number of ways they can be seated, is

• 11! 10!

• $\frac{11!}{6! 5!}$

• $\frac{10! 9!}{5!}$

• $\frac{11! 10!}{5!}$

If t_n denotes the number of triangles formed with n points in a plane, no three of which are collinear and if t_{n + 1} - t_n= 36_, then n is equal to

• 7

• 8

• 9

• 10

$$\begin{gathered} \mathrm{The} \mathrm{term} \mathrm{independent} \mathrm{of} \mathrm{x}\left(\mathrm{x} > 0, \mathrm{x} \ne 1\right) \mathrm{in} \mathrm{the} \mathrm{expansion} \mathrm{of} \\ {\left[\frac{\left(\mathrm{x} + 1\right)}{{\mathrm{x}}^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}} - {\mathrm{x}}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}} + 1} - \frac{\left(\mathrm{x} - 1\right)}{\left(\mathrm{x} - \sqrt{\mathrm{x}}\right)}\right]}^{10} \mathrm{is} \end{gathered}$$

• 105

• 210

• 315

• 420

Out of thirty points in a plane, eight of them are collinear. The number of straight lines that can be formed by joining these points, is

• 296

• 540

• 408

• 348

If n is an integer with 0 $\le \mathrm{n} \le$ 11, then the minimum value of n!(11 - 1)! is attained when a value of n equals to

• 11

• 5

• 7

• 9