Permutations an...

$$\dfrac { { C }_{ 1 } }{ { C }_{ 0 } } +\dfrac { 2{ C }_{ 2 } }{ { C }_{ 1 } } +\dfrac { { 3C }_{ 3 } }{ { C }_{ 2 } } +.....+\dfrac { { nC }_{ n } }{ { C }_{ n-1 } } =$$

The number of odd numbers lying between 40000 and 70000 that can be made from the digits 0, 1, 2, 4, 5, 7 if digits can be repeated any number of times is 

If $${a_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}},} $$ then $${a_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}},} $$ equals 

$${  }^{ n-2 }{ C }_{ r }+2{  }^{ n-2 }{ C }_{ r-2 }+{  }^{ n-2 }{ C }_{ r-2 }$$ equals:

Total number of 6-digit numbers in which all the odd digits and only odd digits appears, is

$$\sum\limits_{r = 1}^n {\frac{{r{.^n}{C_r}}}{{^n{C_{r - 1}}}}} $$ is equal to:

If $$^{2n}C_3$$ : $$^{n}C_2$$ : : $$44 : 1$$, then the value of n is 

$$\sum _{ r=0 }^{ n-1 }{ \frac { {  }^{ n }{ C }_{ r } }{ {  }^{ n }{ C }_{ r }+{  }^{ n }{ C }_{ r+1 } }  } $$ is equal to :

The value of $$^{50}{C_4} + \sum\limits_{r = 1}^6 {^{56 - r}{c_3}} $$ is 

If $$_{  }^{ n-1 }{ { C }_{ r } }=({ k }^{ 2 }-3)_{  }^{ n }{ { C }_{ r+1 } }$$, then k belongs to :