$\text{ All }4\text{ digit numbers of the form }{x}_1{x}_2{x}_3{x}_4\text{ are formed by using digits }$$1,2,3,4,5,6,7,8,9\text{ such that }{x}_1\le x_2\le x_3\le x_4\text{ are written in ascending order, If the }$$\text{ number with rank }460\text{ is abcd. Then }\frac{a+b+c+d}{4}=$

  $For x_1=1 then x_2\le x_3\le x_4 from 1,2,.....9$

$$\begin{gathered} possible cases \\ \left(1\right) x_2<x_3<x_4 \Rightarrow number of ways =c\left(9,3\right) \\ and \left(2\right) x_2\le x_3<x_4 or x_2<x_3\le x_4 \end{gathered}$$

$\Rightarrow select two numbers from available 9 numbers$ $and repeat one from selected$

    $\Rightarrow this can be done in 2·C\left(9,2\right) ways$

    

$$\begin{gathered} \left(3\right) x_2=x_3=x_4 \Rightarrow C\left(9,1\right) \\ therefore number of numbers starting with 1 is {}^{9}c_3+{}^{9}c_2+{}^{9}c_2+{}^{9}c_1 \end{gathered}$$

          $= {}^{10}c_3+{}^{10}c_2={}^{11}c_3 \left[ \sin ce {}^{n}c_r+{}^{n}c_{r-1}={}^{n+1}c_r\right]$

 $$\begin{gathered} similarly number of numbers starting with 2 is {}^{10}c_3=120 \\ number of numbers starting with 3 is {}^{9}c_3=84 \end{gathered}$$

$$\begin{gathered} number of numbers starting with 4 is {}^{8}c_3=56 \\ number of numbers starting with 5 is 7c_3=35 \end{gathered}$$

    $$\begin{gathered} so 460 th number = height number with 5 =5999=abcd \\ now\frac{ a+b+c+d}{4}=\frac{32}{4}=8 \end{gathered}$$