The number of solutions of the equation $\cos 2\theta \cos \frac{\theta }{2}+\cos \frac{5\theta }{2}=2{\cos }^3\frac{5\theta }{2}$ in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ is :

If $\theta ϵ\left[-\frac{7\pi }{6},\frac{4\pi }{3}\right]$, then the number of solutions of $\sqrt{3}{\mathrm{cosec}}^2\theta -2(\sqrt{3}-1)\mathrm{cosec}\theta -4=0$, is equal to :

The number of solutions of the equation $(4-\sqrt{3})\sin x-2\sqrt{3}{\cos }^{2}x=-\frac{4}{1+\sqrt{3}},x\in \left[-2\pi ,\frac{5\pi }{2}\right]$ is

Let $ P $ be the set of seven digit numbers with sum of their digits equal to 11. If the numbers in $ P $ are formed by using the digits 1, 2 and 3 only, then the number of elements in the set $ P $ is :

If all the words with or without meaning made using all the letters of the word "KANPUR" are arranged as in a dictionary, then the word at 440th position in this arrangement is :

From all the English alphabets, five letters are chosen and are arranged in alphabetical order. The total number of ways, in which the middle letter is ' M ', is :

In a group of 3 girls and 4 boys, there are two boys $B_1$ and $B_2$. The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but $B_1$ and $B_2$ are not adjacent to each other, is :

The number of words, which can be formed using all the letters of the word "DAUGHTER", so that all the vowels never come together, is :

Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group $A$ and the remaining 3 from group $B$, is equal to :

Let ${}^n{C}_{r-1}=28,{}^n{C}_r=56$ and ${}^n{C}_{r+1}=70$. Let $A(4\mathrm{cost},4\sin t),B(2\sin t,-2\cos t)$ and $C\left(3r-n,r^2-n-1\right)$ be the vertices of a triangle $A B C$, where $t$ is a parameter. If $(3x-1{)}^2+(3y{)}^2$ $=\alpha$, is the locus of the centroid of triangle ABC , then $\alpha$ equals

The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0 , $1,2,3,4,5,6,7$, such that the sum of their first and last digits should not be more than 8 , is

There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is

The number of sequences of ten terms, whose terms are either 0 or 1 or 2 , that contain exactly five 1 s and exactly three 2 s , is equal to :

From a group of 7 batsmen and 6 bowlers, 10 players are to be chosen for a team, which should include atleast 4 batsmen and atleast 4 bowlers. One batsmen and one bowler who are captain and vice-captain respectively of the team should be included. Then the total number of ways such a selection can be made, is

The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is :

Line $L_1$ of slope 2 and line $L_2$ of slope $\frac{1}{2}$ intersect at the origin O . In the first quadrant, ${\mathrm{P}}_1$, $P_2,\dots ,P_{12}$ are 12 points on line $L_1$ and $Q_1,Q_2,\dots ,Q_9$ are 9 points on line $L_2$. Then the total number of triangles, that can be formed having vertices at three of the 22 points $\mathrm{O},{\mathrm{P}}_1,{\mathrm{P}}_2,\dots ,{\mathrm{P}}_{12}$, ${\mathrm{Q}}_1,{\mathrm{Q}}_2,\dots ,{\mathrm{Q}}_9$, is:

The number of solutions of the equation $\left(\frac{9}{x}-\frac{9}{\sqrt{x}}+2\right)\left(\frac{2}{x}-\frac{7}{\sqrt{x}}+3\right)=0$ is :

If the set of all $a\in \mathbf{R}$, for which the equation $2x^2+(a-5)x+15=3a$ has no real root, is the interval ( $\alpha ,\beta$ ), and $X=|x\in Z;\alpha <x<\beta |$, then $\sum _{x\in X}{x}^2$ is equal to:

Let ${\alpha }_{\theta }$ and ${\beta }_{\theta }$ be the distinct roots of $2x^2+(\cos \theta )x-1=0,\theta \in (0,2\pi )$. If m and M are the minimum and the maximum values of ${\alpha }_{\theta }^4+{\beta }_{\theta }^4$, then $16(M+m)$ equals :