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I will clarify to you the problem approach.I guess you didn't understand it clearly.
The no. of possible such 3-digit numbers are 3! i.e. 6 numbers.

Now,each no. can only occur for 2 times in units/tens/hundreds place.
so, for units place = 2x(3+4+5) = 24
for tens place= 2x10(3+4+5)=240
for hundreds place = 2x100(3+4+5)=2400

Hence,the sum of the digits will be 2400+240+24 = 2664.



Alternatively,since only 6 numbers are there you can even write it down and solve the question
345
354
435
453
534
543

Quickly,add the no.s and you will get the result.

But if your problem approach is clear,I found the formal method to quicker in this question.

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Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker

see here we have three numbers 3,4,5 now if we take 1no. say 3 we have 2 left nos. i.e. 4 & 5 which can be placed in two ways as 45 and 54
similarly taking 4 and 5 at hundreds place , you can easily observe each number is going to repeat 2 times at 100's 10's and unit place
so the sum of all numbers which can be formed= (3+4+5)x2x100+ (3+4+5)x2x10+ (3+4+5)x2x1 =24 x(100+10+1) = 24x111 = 2664.

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Nothing is Impossible, even Impossible says I'm possible.

What is the sum of all possible 3-digit numbers that can be constructed using the digits 3,4,5, if each digit can be used only once in each number?

Here's how I solved it:

Number of integers: 543-345/1+1=199
Average: 345+543/2=444
Sum= 444*199=88,356

What's the correct answer?

You know that each digit will be in each place twice because of the combinations left for the other numbers.

For example, 3 will be in the hundreds place twice because the digits 4 & 5 can only form two different numbers: 45 & 54. Hence the two numbers that have 3 in the hundreds column are 345 & 354.

This example can be extended to the other digits and other place values.

As a result, (3+3+4+4+5+5)*100 + (3+3+4+4+5+5)*10 + (3+3+4+4+5+5)*1 = 24*100 + 24*10 + 24*1, which is what everyone else has typed up.

Knowing how to set up the question is the more difficult part of solving a problem, not the math. This is especially true with the more difficult quantitative questions imo.

If I want to follow the strategy I used above, what is it that I am doing wrong?

The average of all possible numbers will not always be the average of the minimum and maximum number.

For example, 1, 2, and 5.

Min: 125
Max: 521
Average of the Min and Max: 323



List of Numbers:
125
152
215
251
512
521

Average of the List of Numbers:296


I believe the method using the min and max only worked in the original poster's case because the absolute difference of the 1st and 3rd digit from the median digit are equal, i.e. |4-3| = |4-5|.

You're right, the mistake I did here is that I assume that the integers are consecutive, while they are not..and the theory that average of sequence is equal to average of smallest and biggest only holds true in CONSECUTIVE integers..

Do you have an alternative approach to the one being presented above? I usually like to follow systematic approaches, rather than figure the pattern or think of the "logic" behind it. Any other approaches would greatly be appreciated. Thanks