How many 5 digit numbers exist having exactly two 4s in them?

1) 6804
2) 5427
3) 3888
4) 1377

• the five digit number can be represented as xxxx out of these 4s can appear in any two positions
  zero cant appear in first position so when 4 does not occupy first position ,it can occupy any other four positions.the number of posibilities is=4C2*8*9*1*1=3888
  when 4 appears in the first position four occupy any of the other four positions.
  the number of posibilities is 4C1*9*9*9*1=2916
  total numbers=2916+3888=6804
• 10 years agoHelpfull: Yes(0) No(1)
• 1)6804
  1st case
  first digit from left is 4
  so 4c1*9*9*9
  we know that first palace is 4
  we choose a palace 4c1 and fixed it by 4
  4*9*9*9=2916
  2nd case
  first digit from left is not 4
  so 4c2*8*9*9=6*8*9*9=3888
  so total numbers =2916+3888=6804
  6804 is right answer
• 8 years agoHelpfull: Yes(0) No(0)