Find the no. of all 6-digit natural no.s such that the sum of their digits is 10 and each of the digits 0, 1, 2, 3 occurs at least once in them.
1.490
2.386
3.520
4.620

• Answer is 490
  
  The question says 0,1,2,3 should ATLEAST occur once in the digit, it does not say that they should be the only digits used
  
  The combinations to get sum as 10 are:
  0,1,2,3,2,2
  
  So we get => 5*5*4*3*2*1/(3!) = 100 ways
  (5 in the first place because zero cant be used there)
  
  next- 0,1,2,3,3,1
  So we get => 5*5*4*3*2*1/(2!. 2!) = 150
  
  next- 0,1,2,3,4,0
  So we get => 4*5*4*3*2*1/( 2!) = 240
  (4 is multiplied for the first position because there are two zeroes)
  
  
  100+150+240 = 490 !
  
• 10 years agoHelpfull: Yes(40) No(6)
• the correct answer is 490.
  since the combination of digits having sum 10 are 1,2,3,0,2,2
  or 1,2,3,0,3,1
  or 1,2,3,0,4,0
  finding the permutation for all three sets seperately we get (100+150+240) = 490
• 10 years agoHelpfull: Yes(16) No(6)
• Yes the correct choice from my buk was 490.
  1.Three 2, One 3, One 1, One 0
  (6! / 3!) - (5! / 3!)
  = 5.5! / 3!
  = 100 ways
  
  2) One 2, Two 3, Two 1, One 0
  (6! / 2!.2!) - (5! / 2!.2!)
  = 5.5! / 4
  = 150 ways
  
  3) One 4, One 2, One 3, One 1, Two 0
  (6!/2!) - 5!
  = 240 ways
  
  Total = 100 + 150 + 240
  = 490 such numbers
• 10 years agoHelpfull: Yes(8) No(1)
• @karthika in the questions you didn't write 4, then why are u using 4 digit also???
• 10 years agoHelpfull: Yes(4) No(1)
• i m not saying that '0' occurs only at those places. it can occur at any place other than first place.
  u can write the sets like 3,2,1,2,2,0
  or 1,2,3,3,1,0
  or 1,2,3,4,0,0
  what matters is the choice of digits and its combination.
• 10 years agoHelpfull: Yes(3) No(0)
• DIVYA SRIVASTAVA ,the question is all about forming 6 digit numbers using four digits {0,1,2,3},but u used 4.howis it possible
• 10 years agoHelpfull: Yes(3) No(0)
• Answer is 490
  
  The question says 0,1,2,3 should ATLEAST occur once in the digit, it does not say that they should be the only digits used
  
  The combinations to get sum as 10 are:
  0,1,2,3,2,2
  
  So we get => 5*5*4*3*2*1/(3!) = 100 ways
  (5 in the first place because zero cant be used there)
  
  next- 0,1,2,3,3,1
  So we get => 5*5*4*3*2*1/(2!. 2!) = 150
  
  next- 0,1,2,3,4,0
  So we get => 4*5*4*3*2*1/( 2!) = 240
  (4 is multiplied for the first position because there are two zeroes)
  
  
  100+150+240 = 490 !
• 10 years agoHelpfull: Yes(3) No(1)
• no are lyk=321040=360
  =321031=180
  =321022=120
  which i greater than 620 so not sure is this r8 or wrong option??
• 10 years agoHelpfull: Yes(1) No(3)
• as the sum is 10 & 1,2,3,0 are must so the remaning 4, can be as 2,2/3,1/4,0 & in a 6 digit if 0 comes frt it wll becm 5 digt so 386
• 10 years agoHelpfull: Yes(1) No(1)
• kartika can u xplain the solution once again clearly????
• 10 years agoHelpfull: Yes(1) No(0)
• how can u say that '0' occurs only at those places??
• 10 years agoHelpfull: Yes(0) No(0)