What is the sum of all 5 digit numbers which can be formed with the digits 0,1,2,3,4 without repetation?

a)2599980
b)235500
c)923580
d)765432

• 10(11111)*4! - 10(1111)3! = 2599980
• 9 years agoHelpfull: Yes(29) No(6)
• There are 4*4*3*2=96 such numbers. Leading digits can be 1,2,3 and 4 [24 occurrences each] . Other 72 numbers will 0's and hence 72*4-72=216 total occurrences of 1,2,3 and 4 in other places. 216/4=54 1's, 2's, 3's and 4's. [18 1's,2's,3's and 4's at hundredth, tenth and unit place.]
  [1+2+3+4]*24*1000[occurrences at thousand's place]+[1+2+3+4]*18*[1+10+100]=259980
• 9 years agoHelpfull: Yes(8) No(2)
• The formula is (n-1)![Sum of digits][No. of 1s]
  n=No. of digits used
  (5-1)![0+1+2+3+4][11111]=2666640
• 9 years agoHelpfull: Yes(8) No(3)
• The best solution approved here is not correct,the correct answer is option (a) 2599980,checked from two sources...
• 9 years agoHelpfull: Yes(6) No(3)
• ans is 923580 . it is because if we consider 1 as unit digit there are 3*3*2*1 cases so unit digit and similarly considering for 2,3,4,0 unit digit will be 18*1+18*2+18*3+18*4=180 , 18 will be the carry for ten's digit also it will be 180+18=198 ten's digit will be 8 and in options only c option has ten's digit 8.
• 9 years agoHelpfull: Yes(5) No(45)
• 4*4!=96 ,five digit numbers are possible.at unit place,10th place, 100 place and 1000 place each digit is repeating 18 times but in 10000 place each digit is repeating 24 times(excluding zero)
  hence, 18*1+18*2+18*3+18*4=180 so,unit digit will be 0 and 18 will be the carry for 10th place. similarly for 10th place 180+18=198, 8 will be 10th pace digit 19 will be the carry for 100 place.
  in 10000 place 24*1+24*2+24*3+24*4=240 and 240+19(carry from 1000 place)=259
  hence the ans. will be 2599980
  
• 9 years agoHelpfull: Yes(3) No(0)
• There are 4⋅4⋅3⋅2=96 possible numbers (not 5⋅4⋅3⋅2, since the first digit must by nonzero). If you pick one at random, the expected value of the first digit is 1+2+3+44=5/2, and the expected value of each other digit must be 15/8, because the sum of all five digits is 0+1+2+3+4=5/2+4⋅15/8. By linearity of expectation, the sum of all 96 numbers is 96(5/2⋅1000+15/8⋅111)=259980.
• 9 years agoHelpfull: Yes(2) No(2)
• its a 5 digit number therefore (10000+1000+100+10+1)=11111
  (0+1+2+3+4) (11111) (4!) - (0+1+2+3+4) (1111) (3!)
  2599980
• 7 years agoHelpfull: Yes(2) No(0)
• Refer solutions on: http://www.quora.com/What-is-the-sum-of-all-4-digit-numbers-that-can-be-formed-by-the-digits-0-1-2-3-4-No-repetitions-allowed
  
  There are 4⋅4⋅3⋅2=96 possible numbers (not 5⋅4⋅3⋅2, since the first digit must by nonzero). If you pick one at random, the expected value of the first digit is 1+2+3+44=52, and the expected value of each other digit must be 158, because the sum of all five digits is 0+1+2+3+4=52+4⋅158. By linearity of expectation, the sum of all 96 numbers is 96(52⋅1000+158⋅111)=259980.
• 9 years agoHelpfull: Yes(1) No(1)
• lol !!! best solution approved here is actually wrong... RIP question poster :p
• 9 years agoHelpfull: Yes(1) No(0)
• (o+1+2+3+4+5)(11111)(4!)
• 9 years agoHelpfull: Yes(0) No(6)
• why 11111 is multiplied
  
• 9 years agoHelpfull: Yes(0) No(2)
• what is the logic of multiplying 10 please explain @ANKIT PATEL
• 9 years agoHelpfull: Yes(0) No(1)
• sorry m not understanding... plzz elaborate me ans with sol.
  
• 9 years agoHelpfull: Yes(0) No(1)
• Right answer is A
  
  
• 9 years agoHelpfull: Yes(0) No(0)
• 2599980 is right answer
• 9 years agoHelpfull: Yes(0) No(0)