The number of 5-digit numbers that can be formed from 4,2,0,7 and 3 are:
A) 120
B) 96
C) 72
D) 24

• ans-96
  explanation-first place can be filled by 4 ways,similarly 2nd by 4 , 3rd by 3 4th by 2 and 5th by 1.
• 11 years agoHelpfull: Yes(25) No(0)
• 4*4*3*2*1=96;
• 11 years agoHelpfull: Yes(9) No(0)
• since the question contains a zero(0)
  
  so 1st place can't be taken by zero so.
  for 4 other no.ie.
  4,2,3,7.
  each no. can form 5 dgit no.
  so
  let a number starts from 4.then,
  so after 4 the left spaces can be filled in 4! ways
  and same for all 3 digits left..
  so 4!+4!+4!+4!=96 ans
  
• 11 years agoHelpfull: Yes(7) No(7)
• 5!=120
  but when 0 lies in first position, only 4 digits are count.
  hence there are 4!(24) chances of 0 lying in first position.
  
  Hence the answer is 5!-4!=120-24=96 is the answer.
• 11 years agoHelpfull: Yes(5) No(3)
• most significant location takes (4,2,7,3)in 4 ways.remaining four locations take 4! ways(repeatations not allowed=4*4!=96
• 11 years agoHelpfull: Yes(4) No(0)
• As they did not menction about the repetition,
  4*4*3*2*1=96 (for 5 digit number 0 will not be in 1st position)
  
• 11 years agoHelpfull: Yes(1) No(0)
• 5 distinct digits can be permuted in 5! ways .
  the 1st digit can be taken in 5 possible ways(0/2/3/4/7),
  the 2nd digit can be taken in 4 possible ways since repetition is not allowed,
  similarly,the 3rd digit can be taken in 3 possible ways(left over digits),
  4th digit can be taken in 2 possible ways,
  5th digit can be taken in only 1 way since all the other digits were choosen for the previous positions.
  Therefore,the total number of digits that can be formed from 4,2,0,7,3 are
  5*4*3*2*1=5!=120
  
• 8 years agoHelpfull: Yes(0) No(5)
• The number of 5-digit numbers that can be formed from 4,2,0,7 and 3
• 4 years agoHelpfull: Yes(0) No(0)
• If we take 1st number as 0 it becomes four digits number soo if we take remaining 4 numbers are taken in a 1st position and we can make 24 numbers with remaining digits soo total 24+24+24+24=96
• 3 Months agoHelpfull: Yes(0) No(0)