Permutations an...

The value of $$\sum^{10}_{r=0}\begin{pmatrix}10\\r\end{pmatrix}\begin{pmatrix}15\\14-r\end{pmatrix}$$ is equal to

Two lines intersect at $$O$$. Points $$A_{i}$$ and $$B_{i} (i = 1, 2, ...., n)$$ are taken on these two lines respectively, the number of triangles that can be drawn with the help of these $$2n + 1$$ points is

Value of $$\sum _{ k=0 }^{ n }{ { _{  }^{ k }{ C } }_{ n }\sin { \left( kx \right)  } \cos { \left( n-k \right)  }  } $$ is

In a triangle $$ABC$$, the value of the expression $$\displaystyle \sum_{r = 0}^{n}\ ^{n}C_{r}a^{r}.b^{n - r}.\cos (rB - (n - r)A)$$ is equal to

Seven person $$P_1,P_2......, P_7$$ initially seated at chairs $$C_1,C_2,.....C_7$$ respectively.They all left there chairs simultaneously for hand wash. Now in how many ways they can again take seats such that no one sits on his own seat and $$P_1$$, sits on $$C_2$$ and $$P_2$$ sits on $$C_3$$ ?

If $$m$$ denotes the number of $$5$$ digit numbers if each successive digits are in their descending order of magnitude and $$n$$ is the corresponding figure. When the digits and in their ascending order of magnitude then $$(m-n)$$ has the value

There are $$2$$ identical white balls, $$3$$ identical red balls and $$4$$ green balls of different shades. The number of ways in which they can be arranged in a row so that atleast one ball is separated from the balls of the same colour, is

Two classrooms A and B having capacity of $$25$$ and $$(n-25)$$ seats respectively. $$A_n$$ denotes the number of possible seating arrangements of room $$'A'$$, when 'n' students are to be seated in these rooms, starting from room $$'A'$$ which is to be filled up to its capacity. If $$A_n-A_{n-1}=25!(^{49}C_{25})$$ then 'n' equals:

If $$\displaystyle \sum_{k = 1}^{n = 1} \ ^{n - k}C_{r} = ^{x}C_{y}$$ then

If $${ \left( 1+x \right)  }^{ n }=\sum _{ r=0 }^{ n }{ { a }_{ r }{ x }^{ r } } $$ and $${ b }_{ r }=1+\cfrac { { a }_{ r } }{ { a }_{ r-1 } } $$ and $$\prod _{ r=1 }^{ n }{ { b }_{ r }=\cfrac { { \left( 101 \right)  }^{ 100 } }{ 100! }  } $$, then $$n$$ equals to: