(i) Find the sum of all 4-digit numbers that can be formed using the digits 1, 3, 5, 7, 9

(ii) Find the sum of all 4 digited numbers that can be formed using the digits 1, 2, 4, 5, 6 without repetition.

(iii)Find the sum of all 4 digited numbers that can be formed using the digits 0, 2, 4, 7, 8, without repetition.

(i) Given digits are 1, 3, 5, 7, 9 No. of distinct digits n = 5 The sum of the r – digited numbers that can be
formed using the given n distinct digits$(1\le n\le 9)$ is  ${ }^{n-1}{P}_{r-1}\times$ sum of the digits × 111..1 (r times)
Hence n = 5, r = 4
The sum of all 4 digited numbers that can be formed using the digits {1,3,5,7,9}without repetitions is
${ }^{5-1}{P}_{4-1}\times (1+3+5+7+9)\times 1111$

${=}^4{P}_3\times 25\times 1111$

$=24\times 25\times 1111$

=666600

Given digits are 1, 2, 4, 5, 6 No. of digits n = 5 The sum of all r ^– digited numbers that can be formed using the given n distinct digits $(1\le r\le 9)$ is ${ }^{n-1}{P}_{r-1}\times$ sum of the digits × 1111....1 (r times) Hence n = 5, r = 4 The sum of all 4 – digited numbers that can be  formed using the digits {1,2,4,5,6}without repetition is

${=}^{5-1}{P}_{4-1}\times (1+2+4+5+6)\times 1111$

${=}^4{P}_3\times 18\times 1111$

=$24\times 18\times 1111$

=479952

Given digits are 0, 2, 4, 7, 8 No. of digits n = 5 If zero is one among the given n digits, then the sum of the r − digited numbers that can be formed using the given n distinct digits $(0\le n\le 9)$ is

${ }^{n-1}{P}_{r-1}\times$sum of the digits $\times$1111......1

(r times) -${ }^{n-2}{P}_{r-2}\times$sum of the digits$\times$1111......1 (r– 1)times)

Hence n = 5, r = 4 The sum of all 4 digited numbers that can be formed using the digits {0,2,4,7,8}without repetition is

${ }^{5-1}{P}_{4-1}\times (0+2+4+7+8)\times 1111-$

${ }^{5-2}{P}_{4-2}\times (0+2+4+7+8)\times 111$

${=}^4{P}_3\times 21\times 1111-3P_2\times 21\times 111$

$=24\times 21\times 1111-6\times 21\times 111$        = 5,45,958