JEE Main 2022 — Permutation Combination Question with Solution

Let the three-digit odd number be $xyz$

Now $z=1/3/5/7/9$

$x+y+z=7/14/21$  [for sum of digit to be multiple of $7$]

$x+y =6/4/2/13/11/9/7/5/20/18/16/14/12$

When $x+y=6\Rightarrow \left(1,5\right),\left(2,4\right),\left(3,3\right),\left(4,2\right),\left(5,1\right),\left(6,0\right)$

Total possibilities$=6$

When $x+y=4\Rightarrow \left(1,3\right),\left(2,2\right),\left(3,1\right),\left(4,0\right)$

Total possibilities$=4$

When $x+y=2\Rightarrow \left(1,1\right),\left(2,0\right)$

Total possibilities$=2$

When $x+y=13\Rightarrow \left(4,9\right),\left(5,8\right),\left(6,7\right),\left(7,6\right),\left(8,5\right),\left(9,4\right)$

Total possibilities$=6$

When $x+y=11\Rightarrow \left(2,9\right),\left(3,8\right),\left(4,7\right),\left(5,6\right),\left(6,5\right),\left(6,5\right),\left(7,4\right),\left(8,3\right),\left(9,2\right)$

Total possibilities$=8$

When $x+y=9\Rightarrow \left(1,8\right),\left(2,7\right),\left(3,8\right),\left(4,5\right),\left(5,4\right),\dots .\left(8,1\right),\left(9,0\right)$

Total possibilities$=9$

When $x+y=7\Rightarrow \left(1,8\right),\left(2,5\right),\left(3,4\right),\dots .\left(8,1\right),\left(7,0\right)$

Total possibilities$=7$

When $x+y=5\Rightarrow \left(1,4\right),\left(2,3\right),\left(3,2\right),\left(4,1\right),\left(5,0\right)$

Total possibilities$=5$

When $x+y=20$  No possibilities

When $x+y=18\Rightarrow \left(9,9\right)$

Total possibilities$=1$

When $x+y=16\Rightarrow \left(7,9\right),\left(8,8\right),\left(9,7\right)$

Total possibilities$=3$

When $x+y=14\Rightarrow \left(5,9\right),\left(6,8\right),\left(7,7\right),\left(8,6\right),\left(9,5\right)$

Total possibilities$=5$

When $x+y=12\Rightarrow \left(3,9\right),\left(4,8\right),\left(5,7\right),\left(6,6\right)\dots .\left(9,3\right)$

Total possibilities$=7$

Hence, the total number of three digit odd numbers whose sum of digit is divisible by $7$ is $63$