What is the sum of all 3 digits number that can be formed using digits 0,1,2,3,4,5 with no repitition

Question

any one plz explain it clearly

Answer 1

First we calculate the sum of all possible 3 digit numbers with 6  digits 0,1,2,3,4,5, allowing 0 also to be the first digit.

• There are 6x 5 x 4  = 120 numbers.
• The digits are evenly dispersed, so each digit appears 120/5 = 24 times in the 100's place,10th place and unit place.
• Summing the 100's place numbers, we get 24 * (0+1+2+3+4+5) * 100 = 24*15*100= 36000.
• Summing the 10's place numbers, we get 3600; 10's.
• 1's place, 360.
• Adding together, 39960.

Now we have to subtract the numbers start with 0.

• They have a 0 in the 100's place and a digit from 1 to 5 in each of the other 2 places.
• There are 5x4 = 20 of them.
• Again 5 digits are distributed evenly, each one appearing 20/5 = 4 times
•  so the sum in the 100's place is 4 * (1 + 2 + 3 + 4 +5) * 10 = 600; the sum in the 1's place is 60.
• Adding together, 6600.

Subtracting the second sum from the first sum we get a final answer of 39960 - 6600 = 39300

Comments

but i think u r giving this answer for the 4 digit number but q says about 3

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Oh yea! My bad! Sorry.. answer is modified now!!!

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i think its still wrong   each digit appears 120/5 = 24 times ??? why divide by 5 this should be 120/3 ,

also i didint understand this line  we get 24 * (0+1+2+3+4+5) * 100  when we have to take 3 digits

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Well, there are not $120$ numbers. ( $0$ at MSB doesn't makes a $3$ digit number.)

Total $3$ digit numbers = $5*5*4 = 100$

Brother modify your answer :)