Permutations and Combinations

Find the sum of all the numbers that can be formed with the digits $2, \; 3, \; 4, \; 5$ taken all at a time.

When $5$ is in the unit's place, the remaining 3 digits ($2, \; 3, \; 4$) can be arranged amongst themselves in $3! = 6$ ways.

i.e. $\;$ $6$ numbers are possible with 5 in the unit's place.

$\therefore \;$ Sum of unit's place $= 6 \times \left(5 \times 1\right) = 30$

When $5$ is in the ten's place, the remaining 3 digits ($2, \; 3, \; 4$) can be arranged amongst themselves in $3! = 6$ ways.

i.e. $\;$ $6$ numbers are possible with 5 in the ten's place.

$\therefore \;$ Sum of ten's place $= 6 \times \left(5 \times 10\right) = 300$

When $5$ is in the hundred's place, the remaining 3 digits ($2, \; 3, \; 4$) can be arranged amongst themselves in $3! = 6$ ways.

i.e. $\;$ $6$ numbers are possible with 5 in the hundred's place.

$\therefore \;$ Sum of hundred's place $= 6 \times \left(5 \times 100\right) = 3000$

When $5$ is in the thousand's place, the remaining 3 digits ($2, \; 3, \; 4$) can be arranged amongst themselves in $3! = 6$ ways.

i.e. $\;$ $6$ numbers are possible with 5 in the thousand's place.

$\therefore \;$ Sum of thousand's place $= 6 \times \left(5 \times 1000\right) = 30000$

$\therefore \;$ Sum of all possible numbers with 5 in the unit's, ten's, hundred's and thousand's place is

$= 30 + 300 + 3000 + 30000 = 33330$

Similarly, sum of all possible numbers with 4 in the unit's, ten's, hundred's and thousand's place is

$= 6 \times \left(4 \times 1\right) + 6 \times \left(4 \times 10\right) + 6 \times \left(4 \times 100\right) + 6 \times \left(4 \times 1000\right)$

$= 24 + 240 + 2400 + 24000 = 26664$

Sum of all possible numbers with 3 in the unit's, ten's, hundred's and thousand's place is

$= 6 \times \left(3 \times 1\right) + 6 \times \left(3 \times 10\right) + 6 \times \left(3 \times 100\right) + 6 \times \left(3 \times 1000\right)$

$= 18 + 180 + 1800 + 18000 = 19998$

Sum of all possible numbers with 2 in the unit's, ten's, hundred's and thousand's place is

$= 6 \times \left(2 \times 1\right) + 6 \times \left(2 \times 10\right) + 6 \times \left(2 \times 100\right) + 6 \times \left(2 \times 1000\right)$

$= 12 + 120 + 1200 + 12000 = 13332$

$\therefore \;$ Sum of all the numbers that can be formed with the digits $2, \; 3, \; 4, \; 5$ taken all at a time

$= 33330 + 26664 + 19998 + 13332 = 93324$