Permutations and Combinations

$LHS={\displaystyle \sum _{r=1}^{15}r\left(r!\right)}$

$=\sum \left(r+1-1\right)r!={\displaystyle \sum _{r=1}^{15}\left(r+1\right)!-r!}$

$=\left(2!-1!\right)+\left(3!-2!\right)+....+\left(16!-15!\right)$

$\begin{array}{l}=16!-1\\ {=}^{16}{P}_{16}-1\end{array}$

The 1^st such digit is 11 * 19 = 209

Sum = [209 + 220 + 231 + .+ 495] - [231 + 319 + 341 + 418 + 451] = 7744

${\displaystyle \sum _{r=1}^{20}\left(r^2+1\right)r!}$  

t_r = (r² + 1)r!

= r²r! + r!

= r(r + 1 – 1)r! + r!

= r(r + 1)! – (r – 1)r!

= V_r – V_r-1

  ${\displaystyle \sum _{r=1}^{20}\left(V_r-V_{r-1}\right)}$              

= V₁ – V₀

$+V_2-V_1$

$+V_3-V_2$

$+V_{20}-V_{19}$

$+V_{20}-V_{19}$

$=V_{20}-V_0=20\left(21!\right)-0$

$\left(22-2\right)\left(21!\right)=22!-2\left(21!\right)$        

$a\in \left\{1,2,3,....,9\right\}$  

  $b\in \left\{0,1,2,...,9\right\}$              

9 * 9 = 81

81 + 81 + 81 = 243

${\displaystyle \sum _{r=1}^{20}\left(r^2+1\right)r!}$

t_r = (r² + 1)r!

= r²r! + r!

= r (r + 1 – 1)r! + r!

= r (r + 1)! – (r – 1)r!

= V_r – V_r-1

_{${\displaystyle \sum _{r=1}^{20}\left(V_r-V_{r-1}\right)}$}

= V₁ – V₀

_{$+V_2-V_1$}

_{$+V_3-V_2$}

_{$+V_{20}-V_{19}$}

_{$=V_{20}-V_0=20\left(21!\right)-0$}

_{$\left(22-2\right)\left(21!\right)=22!-2\left(21!\right)$}

Arranging letter in alphabetical order A   D   I   K   M   N   N for finding rank of MANKIND making arrangements of dictionary we get

A …….->

$\frac{6!}{2!}=360$        

D ………->360

l ………. ->360

K ………->360

MAD …….->

$\frac{4!}{2!}=12$        

$\therefore$ Rank of MANKIND = 1440 + 36 + 12 + 2 + 2 = 1492.

1012 $\le$  Number in the question $\le$  23421

Also, the number has to use digits {2, 3, 4, 5, 6} without repetition and the number has to be divisible by $\frac{55}{11*5}$  

As the number has to be divisible by both 5 and 11,

5 → once place

Let us make 4-digit such numbers first:

{2, 3, 4, 6} (digits are not be repeated)

A number is divisible by 11 it difference of sum of its digits at even places and sum of digits at odd place is 0 or multiple of 11.

Required number = Total – no character from {1, 2, 3, 4, 5}

$=\left(10^6-5^6\right)+\left(10^7-5^7\right)+\left(10^8-5^8\right)$

$=5^6\left(2^6*111-31\right)=5^6*\frac{7073}{\alpha }$

= 7073

$x_1+x_2=6,13\to 6+6=12$

$\begin{array}{l}{x}_1+x_2=4,11,18\to 4+8+1=13\\ x_1+x_2=2,9,16\to 2+9+3=14\\ x_1+x_2=7,14\to 7+5=12\\ x_1+x_2=5,12\to 5+7=12\end{array}$

=63