Find the sum of the number of combinations possible by using 1,2,3,4 such that no number is repeated(ex:- 1223,4322 are invalid 1234,4321 are valid)more than once.

• total no. formed = 4!=4*3*2*1=24
  each digit will be used at each positions, 6 times
  sum of digits at unit place = 6*1+6*2+6*3+6*4 = 6*(1+2+3+4)= 60
  this sum will be same for hundred, tens & unit places
  sum of all 4 digit numbers formed using digits 1,2,3,4
  = 1000*60+100*60+10*60+60
  =60*(1000+100+10+1)
  =60*1111
  =66660
  
• 10 years agoHelpfull: Yes(52) No(5)
• there is a short-cut process:
  let no of digit= n;
  then (n-1)!*(111...up to n times)*(sum of digits)= sum of all possible numbers
  eg: (3!)(1111)(10)=66660
• 10 years agoHelpfull: Yes(23) No(0)
• Clearly if a number is fixed at any position eg 1] _ _ _ ,
  it yields 6 possibilities.
  We are required to find out sum of those nos. of combinations using 1 2 3 and 4.
  Sum at each place= 6(1+2+3+4)= 60
  Sum of the number of combinations possible =60*(1+10+100+1000)=66660 (Ans)
  Check now--Sum of all possible 4 digit nos.---->
  1234+1243+1342+1324+1432+1423+2134+2143+2314+2341
  +2413+2431+3142+3124+3214+3241+3412+3421+4123+4132+4213+4231+4321+4312=
  7998+13776+19554+25332=66660
  See,both results are the same!
• 10 years agoHelpfull: Yes(14) No(1)
• Surendhran i think u guessed it in a wrong way.. Question was the sum of the numbers not digits.. Consider 2 different combinations- 1234+2134=3368.. Clearly, it already crossed 240.. More combinations to come.
• 10 years agoHelpfull: Yes(4) No(2)
• Sum of total combination are 24
• 10 years agoHelpfull: Yes(1) No(4)
• sum of digits: (4+3+2+1)=10 & 4! possible permutations occur. in each place each digit is repeated 6 times. so, 10*6=60 is sum at each place. total sum= (60*1000)+(60*100)+(60*10)+60=66660
• 10 years agoHelpfull: Yes(1) No(0)
• Surendhran i think u guessed it in a wrong way.. Question was the sum of the numbers not digits.. Consider 2 different combinations- 1234+2134=3368.. see it already crossed 240.. more combinations to come
• 10 years agoHelpfull: Yes(1) No(0)
• no of combos = 4*3*2*1=24
  6tyms 6*(1+2+3+4)=60
  for 4 places= 60(1000+100+10+1)
  =66660
• 10 years agoHelpfull: Yes(1) No(0)
• This type of sum has particular formulae to memorize....
  (N-1)!*(a1+a2+a3+.....+aN)*((10^n-1)/9)
  =(4-1)!*(1+2+3+4)*((10^4-1)/9)
  =66660
• 10 years agoHelpfull: Yes(1) No(0)
• answer is:66660
  well easy way is we can solve it by sum of series formula i.e. (n/2)(a+l)
  where a is first term and l is last term
  
  now n is total number which 4!=24(by arrangement)
  thus (24/2)(1234+4321)=66660...here 1234 will be first term and 4321 will last term
• 10 years agoHelpfull: Yes(1) No(0)
• 1234=10
  3421=10
  2143=10
  4312=10
  
  
• 10 years agoHelpfull: Yes(0) No(2)
• total combination is 24
  for each combination addition is 10
  so sum of total combination is 240
  ans is 240
• 10 years agoHelpfull: Yes(0) No(5)
• total combination is 4*3*2*1=24...and the sum of the nos for every no is 4+3+2+1=10..so sum=24*10=240
  
• 10 years agoHelpfull: Yes(0) No(1)
• 23 people are there, they are shaking hands together, how many hand shakes possible, if they are in pair of cyclic sequence.
• 10 years agoHelpfull: Yes(0) No(0)
• my ans comes out to be 66438
  
• 10 years agoHelpfull: Yes(0) No(0)
• Use formula
  n=no. of number series given(4 i.e. 1,2,3,4)
  (sum of no.)*(n no of 1)*(n-1)!
  (1+2+3+4)*(1111)*(4-1)! =66660
• 10 years agoHelpfull: Yes(0) No(0)