How many 4 digit no's can be formed using the digits 0-9 with repetition of digits allowed such that 3 appears exactly twice?

• answer is 459
  here we have a 4 digit number so msb or left most digit cannot be 0
  now we take two cases
  
  case 1
  when 3 is in the msb
  so in the remaining 3 places we require one 3 and two numbers from 0 to 9 except 3 i.e. 9 numbers
  so 3 can be put in 3 places in 3 ways and remaining two places can be filled in 9*9 = 81 ways
  so total number of ways are 81*3 = 243 ways
  
  case 2
  here there is no 3 in msb. both 3 are in remaining three places
  it means msb can be filled by numbers from 0 to 9 except 0 and 3 that is in 8 ways
  and the remaining three places can be filled by two 3 and any of the numbers from 0 to 9 except 3 that is 9 ways
  two 3 can be filled in three places in 3 ways
  so number of arrangements are 8*9*3 = 216
  
  so total number of ways is 216+243 = 459
  
  
• 10 years agoHelpfull: Yes(28) No(2)
• Ans-459
  possible ways can be 33xx xx33 3x3x x3x3 x33x 3xx3 where x can be 0-9 except 3 so 9*9=81 for 1 way so similarly 4 six ways 6*81= 486 - the cases where 0 is at first position
• 10 years agoHelpfull: Yes(4) No(2)
• ans is 459
  here two case will arise
  1)3 is present at thousnd's place
  2)3 is nt there
  total no of arranging twice 3 at 4 places=4C2=6
  from thease 6,3 will contain 3 at thousand's place
  so in this case total1=(1*9*9*1)*3=243
  and for rest 3
  total2=(8*9*1*1)*3=216
  so total=216+243=259
• 10 years agoHelpfull: Yes(4) No(0)
• 9*1*1*10+1*1*10*10+9*1*10*1+1*10*10*1=380
• 10 years agoHelpfull: Yes(1) No(8)