Question

Sum of all numbers that can be formed with permutations of 0,1,2,3

Answer 1

Hi there!
Here's the answer:

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SOLUTION :

¶¶¶ STEP-1 :
Find No. of 4 digits that can be formed using 0, 1, 2 & 3 :

^^^ Note that a digit is zero ^^^
^^^ ASSUMING REPLETION NOT ALLOWED ^^^

Assume the No.s take the place in below blanks

_ _ _ _


• The first blank can be filled in 3 ways (1,2,3)
{ As 0 cannot be at first place}

• Second Blank has 3 possibilities
{0, & excluding the digit used in first blank}

• Third blank has 2 possibilities
{Except digit used in first and second blank}

• Fourth Blank can be filled only in 1 way
{Left over digit }

•°• No. of 4-digit No.s that cam be formed using the digits 0,1,2 & 3 is 3×3×2×1 = 18


¶¶¶ STEP-2 :
• Find No. of times each of the digit appears in each place


Now,
Find K
$K =\frac{Total\: No.\: of\: Options}{No. of\: options\: for\: first\: place}$

Each of the Non zero digit will come K times at first place.

Here, 4-digit No.s are considered, First place is thousands place

• Each of the Non - zero digits viz 1, 2, 3 will come at thousand's place $K = \frac{18}{3}= 6$ in the 18 No.s formed using given digits.

• The digit 0 will come 6{=K} at the 2nd place in 18 No.s formed
The Remaining 12 blanks be shared equally by non zero digits(1,2,3)
=> Each of the digits (1,2,3) will come 4 times at the second place.

• For the third and fourth place, the same possible ways as second.
The digit 0 will come 6{=K} times
The Non zero digits (1,2,3) will come 4 times at the third place and fourth place.


¶¶¶ STEP-3:
Find Sum of the No.s at each place to find sum of all digits formed using given digits.

• Sum of No.s at first (1000's) place
= 6(1+2+3) = 6×6 = 36
=> 36 is the sum of No.s at thousands place


• Sum of No.s at second (100's) place
= 6×0 + 4(1+2+3) = 0 + (4×6) = 24
{As 0 comes 6 times and Non zero digits (1,2 3) comes 4 times in the 18 No. formed}
=> 24 is the sum of No.s at hundred's place


Similiarly,

• Sum of the No.s at third (10's) place
= 6×0 + 4(1+2+3) = 0 + (4×6) = 24
=> 24 is the sum of No.s at ten's place


• Sum of the No.s at fourth (1's) place
= 6×0 + 4(1+2+3) = 0 + (4×6) = 24.
=> 24 is the sum of No.s at unit's place

Now,


¶¶¶ STEP - 4 :
Multiply sum of No.s at each place with the weight of the digit. Add up all the results.

36 × 1000 = 36000
24 × 100 = °°°°2400
24 × 10 = °°°°°°°°240
24 × 1 = °°°°°°°°°°°°24
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TOTAL = °°°°°38664


•°• Required sum = $\underline{\underline{38664}}$


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Hope it helps