Find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition.


I will post OA and OE tomorrow.

We can form a total of 4! or 24 numbers.When we add all these numbers ,let us look at the contribution of of the digit 2 to the sum.

When 2 occurs in the thousand place in a particular number,its contribution in the total will be 2000.the number of numbers that can be formed
with 2 in the thousand place is 3! i.e 6 numbers.Hence when 2 is in the thousands place its contribution to the sum is 3! * 2000

Similarly when 2 occurs in the hundreds place, its contribution to the sum is 3! * 200

Similarly when 2 occurs in the tenth place, its contribution to the sum is 3! * 20

Similarly when 2 occurs in the unit place, its contribution to the sum is 3! * 10

The total contribution of 2 to the sum is 3! *(2000+200+20+1)=3!*2222
In a similar manner ,the contribution of 3,4, and 5 to sum will respectively be 3!*3333,3!*4444 and 3!*5555

Hence total sum using the aove four digits 3!*(2222+3333+4444+5555) i.e 3! *(2+3+4+5) * 1111


Now we can generalize the above

If all the possible n digit numbers using n distinct digits are formed ,the sum of all the numbers so formed is equal to
(n-1)! * ( sum of the n digits ) *( 1111...n times)


Consider giving me KUDOS if this is useful.

Along with the question, do post the options.

if you keep the digit 2 at the one's digit. The total numbers that can be formed is 3! = 6
=> 2 occurs at one's digit 6 times. Similarly all the 4 numbers occurs at one's digit 6 times.

Similar all the 4 numbers occurs at all the 4 digits - one's,ten's, hundred's, thousand's- 6 times.

=> total sum = ( 1000+100+10+1 ) *( 2+3+4+5)*6

= 1111*14*6 = 1111*84 = 93324
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Actually there is the direct formula for this kind of problems. Of course it's better to understand the concept, then to memorize the formula but in case someone is interested here it is:

1. Sum of all the numbers which can be formed by using the \(n\) digits without repetition is: (n-1)!*(sum of the digits)*(111…..n times).

2. Sum of all the numbers which can be formed by using the \(n\) digits (repetition being allowed) is: \(n^{n-1}\)*(sum of the digits)*(111…..n times).

Hope it helps.
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Yes understanding is very imp as pointed by bunuel.

The question becomes interesting if total sum of all the 3 digits number is asked.
Then the give formula is modified.

the 111...n becomes 111....m where m = 3 for 3 digit number.
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Do I get it?
When you have to sum all the 3 digit numbers out of 2,3,4,5 than the answer is:

without repetition:
(3-1)!*14*111= 3108

with repetition:
(3^2)*14*111=13986

How many options do I have? 4*3*2*1
How many digits will be repeated equally in the options? 4

\(=\frac{4*3*2*1}{4}*14*1111=84*1111=64*1111=93,324\)

Hi Bunuel,
So in the below question if we apply the formula shouldn't we consider 4 as n?
Find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition.
(4-1)*(2+3+4+5)*1111
Many thanks for your help.
Aybige

Correct, but you've missed factorial (!). It should be: (4-1)!*(2+3+4+5)*(1111)=93324.
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