how many five digit numbers are there such that two left most digits are even and remaining are odd and 4 digit should not be repeated.

• N=(4*5-1)*5*5*5=2375
  
  where
  4 cases of first digit {2,4,6,8}
  5 cases of second digit {0,2,4,6,8}
  1 case of 44 for two leftmost digit
  5 cases of third digit {1,3,5,7,9}
  5 cases of fourth digit {1,3,5,7,9}
  5 cases of fifth digit {1,3,5,7,9}
• 10 years agoHelpfull: Yes(33) No(10)
• even numbers are 2,4,6,8 and odd 1,3,5,7,9
  
  so choice for first 2 digit 4 and 4
  
  and for last 3 digit 5 ,4 ,3 coz not repeated
  
  so ans 4*4*5*4*3 = 960
• 10 years agoHelpfull: Yes(32) No(16)
• Starting from the left most digit...
  
  the possible combinations for first digit the possible ways are {2,4,6,8}=4 ways. (Here '0' is not included coz it won't make it a 5 digit number.)
  
  Now for second digit total ways are (0,2,4,6,8) =5 ways.
  
  but there is an exception that the digit '4' should not be repeated. so out of total cases i.e 4x5=20 , one such case exists and i.e. 44 remove it. so now the total ways will be 20-1 = 19 ways.
  
  and for 3rd digit possible ways are {1,3,5,7,9} = 5 ways
  
  similarly for 4th digit possible ways are {1,3,5,7,9} = 5 ways
  
  similarly for 5th digit possible ways are {1,3,5,7,9} = 5 ways
  
  hence the total number of ways are 19x5x5x5= 2375
• 8 years agoHelpfull: Yes(16) No(1)
• for 1-st digit total case 2,4,6,8
  for 2-nd digit total case 0,2,4,6,8
  repeat nos are 22,44,66,88
  for 3-rd, 4-th, 5-th digit total case 1,3,5,7,9 but not repeated
  
  so total case (4*5-4)*5*4*3=960
• 10 years agoHelpfull: Yes(9) No(2)
• 4*4*5*4*5=1600
• 10 years agoHelpfull: Yes(8) No(9)
• can anybody giv the correct explanation
• 10 years agoHelpfull: Yes(3) No(2)
• may be 960 numbers are there
• 10 years agoHelpfull: Yes(2) No(4)
• pls give me the correct solution.
• 10 years agoHelpfull: Yes(2) No(2)
• even numbers are 0,2,4,6,8 and odd numbers are 1,3,5,7,9
  so the left most digit can be any of {2,4,6,8) i,e. in 4 ways and second left digit can be any of {0,2,4,6,8} i.e. in 5 ways but as the case 44 is not permitted,hence two left digits can be chosen in (4*5)-1=19 ways
  now remaining 3 numbers can be chosen in 5*5*5 ways(here repetition is allowed)
  so, total case=19*125=2375
• 7 years agoHelpfull: Yes(2) No(0)
• The first two slots need to be even numbers. The first one can be filled by 4 numbers (2, 4, 6, 8). The second can be filled by 4 numbers (0, 2, 6, 8) i.e., exclude 4 and include 0 in the second spot.
  
  Hence, the first two can be filled up in 4 x 4 = 16 ways.
  
  The remaining three slots have to be filled up using any of 1, 3, 5, 7, 9 digits as they can be repeated. i.e.,
  
  third slot filled by 5 digits
  fourth by 5
  and fifth by 5
  
  So, 5 x 5 x 5 = 125 ways of filling the last three slots.
  
  Combining, we have 125 x 16 = 2000.
  
• 8 years agoHelpfull: Yes(1) No(2)
• acc to me
  even numbers are 2,4,6,8 and odd 1,3,5,7,9
  1st 2 no even and next 3 odd
  1st repeat 1 even then repeat 1 odd
  so 4*4*5*4*3 + 4*3*5*4*4 = 1920
• 10 years agoHelpfull: Yes(0) No(2)
• 
  even : 0,2,4,6,8
  odd: 1,3,5,7,9
  
  we cant take zero on left most size as this will not satisfy condition(5 digit num)
  so we have 4 for first digit
  5 for second left most
  and 5 for 3rd digit
  and we cant take digit selected on 3rd left most as it is given that 4th digit should nt repeat.
  hence we have 4 choices for 4th digit and for final digit we can take 4 number excluding num selected on 4th place
  hence ans should be
  4*5*5*4*4 = 1600
• 10 years agoHelpfull: Yes(0) No(2)
• if 4 is not used then 3*4*5*5*5 = 1500 and if 4 used then its 1*4*5*5*5 + 3*1*5*5*5 = 875 then total 1500 + 875 = 2375
• 9 years agoHelpfull: Yes(0) No(0)
• samjh nahin aaya....????
• 9 years agoHelpfull: Yes(0) No(0)
• 4*4*5*5*5=2000.since at first position we can use only 2,4,6,8 not 0 but at second place we can use 0,2,6,8 not 4 because it should not be repeated . and at last 3 place we can arrange simply as 5*5*5 because there are 5 odd digit which can be repeated.
• 8 years agoHelpfull: Yes(0) No(0)