what is the sum of all 5 digits numbers which can be formed with the digits 0,1,2,3,4 without repetition

• There is a simplest formula for solving these type of questions when we are using numbers along with 0 for finding sum.
  i,e........(sum of numbers)*{(n-1)!*(111...........n times)-(n-2)!*(111..........(n-1)times) }
  so,here by putting the values as,
  
  (0+1+2+3+4){(5-1)!*(11111)-(5-2)!*1111}
  
  = 10{(4!*11111)-(3!*1111)}
  
  = 10{266664-6666}
  
  = 2599980 ans
• 8 years agoHelpfull: Yes(66) No(5)
• If u observe each digit will be present at each position in 4! time.
  i.e., No of times 0 is in first(ten thousand) place is (5-1)! = 4!= 24 ways.
  In the same way ,
  No of times 1 is in first place is (5-1)! = 4!= 24 ways.
  No of times 2 is in first place is (5-1)! = 4!= 24 ways.
  No of times 3 is in first place is (5-1)! = 4!= 24 ways.
  No of times 4 is in first place is (5-1)! = 4!= 24 ways.
  No of times 0 is in second( thousand) place is (5-1)! = 4!= 24 ways.
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  No of times 4 in fifth(One's) place is (5-1)! = 4!= 24 ways.
  
  So the sum of all numbers will be
  = (5-1)! * (10000+1000+100+10+1)*(0+1+2+3+4)
  =4! * (11111)*10
  = 24* 11111*10
  = 2666640
  
  Logic can be (n-1)!*(111...1 ntimes)*(sum of digits)
  
  
  As numbers formed by digits 1,2,3,4 and 0 as first digits are not 5 digit numbers, so subtracting 4 digit numbers formed by 1,2,3,4
  i.e., 2666640 - 3!*(1111)*(1+2+3+4)
  = 2666640 - 6666*10
  = 2666640 -66660
  = 2599980
• 8 years agoHelpfull: Yes(18) No(1)
• no bhavya if 0 is included excluded then formulae goes as
  .(sum of numbers)*{(n-1)!*(111...........n times) only ,u r formulae is for the Q like if the digit is given 1,2,3,4,5 and we have to make find sum of number form of 2 digit from these numbers. then u r formulae is applicable. see once. visit this link:-https://www.youtube.com/watch?v=leBWTcqACkg
  
• 8 years agoHelpfull: Yes(8) No(3)
• total no. of 5 digit numbers formed by 0,1,2,3,4 are 4*4!=96 .because at first place 0 cant be placed .
  so,numbers formed when 1 at first place=4!
  numbers formed when 2 at first place=4!
  numbers formed when 3 at first place-4!
  numbers formed when 4 at first place=4!
  total =4*4!=96.
  now sum of all the numbers ,
  at 10000th place each number will come 24 times except 0
  at 1000th place each number will come 24 times
  and so on ...........
  at 10000th place =24*(1+2+3+4)=24*10 =2,40,00,00 (at 10,000th place)
  at 1000th place=24*(0+1+2+3+4+5)=24 *10=2,40,000
  at 100th place=24*(0+1+2+3+4+5)=24*10=2,40,00
  at 10th place=24*(0+1+2+3+4+5)=24*10=2400
  at unit place =24*(0+1+2+3+4+5)=240
  so sum of the numbers =2400000+240000+24000+2400+240=2666640
  
  
• 8 years agoHelpfull: Yes(1) No(5)
• If 0 is not included in the question then,
  
  sum of the given numbers*P(n-1,r-1)(1111.........n times)
  
  where n= nos. of the digits from which we have to find sum,
  
  r= sum of the given digits.
• 8 years agoHelpfull: Yes(1) No(6)
• If 0 is present in given digits then formula is :
  (sum of numbers)*{(n-1)!*(111...........n times)-(n-2)!*(111..........(n-1)times) }..
  and if 0 is not present in given digits then formula is:
  (sum of numbers)*{(n-1)!*(111...........n times)..
  In this case,since 0 is present in the given digits so we will be using (sum of numbers)*{(n-1)!*(111...........n times)-(n-2)!*(111..........(n-1)times) }..
  and solution is:
  (0+1+2+3+4)*(5-1)!*(11111)-(1+2+3+4)*(5-2)!*(1111)
  
  =10*4!*11111-10*3!*1111
  
  =2666640-66660
  
  =2599980
• 4 years agoHelpfull: Yes(1) No(0)
• There are total 96 numbers. Sum of single number obviously be 0+1+2+3+4=10.
  Required Sum is 96*10=960.
• 4 years agoHelpfull: Yes(1) No(0)
• bhavya if 0 is exluded then what will be the method to solve this.
• 8 years agoHelpfull: Yes(0) No(0)
• (0+1+2+3+4)*(4!)(11111)
• 7 years agoHelpfull: Yes(0) No(0)
• Consider the string 123456789. Observe that each of the 99 digits in the string can safely be either included or not included, since failing to include a digit will not violate the increasing order of the digits. Hence, there are 2^9 possible strings that contain (0 or 1 or 2 or ... or 9) digits. However, we likely want to omit the case where there are 00 digits, which leaves us with a final answer of:
  2^9−1=511
  :)
  
• 7 years agoHelpfull: Yes(0) No(0)