how many ways are there to form a 6 digit number using the digits 0-6 such that last 2 places occupt even number.repetition is not allowed

• the last digit can be filled up in 4 ways and the last but one digit can be filled up in 3 ways.so total 4x3=12 ways.
  Now there are 5 digits left but the extreme left digit can be filled up in 4 ways as 0 cannot be used in the extreme left digit.the next can be filled up in remaining (7-3)=4 ways then 3 ways then 2 ways
  so total=4x4x3x2x4x3=1152 ways.
• 11 years agoHelpfull: Yes(29) No(9)
• _ _ _ _ _ _ there are six places to be filled by 7 numbers i.e 0,1,2,3,4,5,6 we have to fill the last two places with only an even number. we have 3 even numbers, 2,4,6 last place can be filled in 3 ways. the second last place place can be filled in 2 ways as we have already used one of the 3 even digits available for the last place. the first digit cannot be filled with 0, hence we can fill it with either 1,3,5 and one of the even number left after filling the last two places. this can be done in 4 ways. similarly 2nd place can be filled with the remainig digits in 4 ways, note that 0 can be included now. and finally 3rd and 4th place can be filled in 3 and 2 ways respectively. hence total no.of ways are 4*4*3*2*2*3= 576
• 11 years agoHelpfull: Yes(22) No(7)
• it will be 576
  4*4*3*2*3*2(first select last two even digit)
• 11 years agoHelpfull: Yes(5) No(2)
• if 0 occupies place in last two digitts
  the possuble out comes are
  0 2
  04
  06
  20
  40
  60
  then we can arrange given 5 numbers in 4 places as 5*4*3*2
  
  
  
  when 0 is not selected in last two places
  
  then last two digits are filled by 2 4 6
  in 3*2 ways
  
  and rest 4 places fillesd by excludiong zero in first place as 4*4*3*2
  
  
  
  so answer is 720+576=1296
• 11 years agoHelpfull: Yes(4) No(0)
• 4*4*3*2*3*2=576
• 11 years agoHelpfull: Yes(3) No(2)
• Alternative Solution:
  Assuming the question says only even digits should occupy the last two positions, the last and last but one places should be filled with 0,2,4,6.
  Thus ways to fill last two places = 4*3 = 12.
  Out of the remaining 5 digits, 0 cannot be in the initial position.
  Hence required number of ways = 12*4*4*3*2 = 1152 ways
• 11 years agoHelpfull: Yes(3) No(1)
• 1c1*2c1*3c1*3c1*4c1=108
• 11 years agoHelpfull: Yes(1) No(6)
• 1056 ways
  
• 11 years agoHelpfull: Yes(0) No(4)
• 0 is also an even number please consider it too.
  The answer would be 432. Please give the correct answer to verify.
• 11 years agoHelpfull: Yes(0) No(2)
• Ans: 2400 ways
  Last two places will be even numbers when we have 0,2,4,6 in the last position.
  PLEASE NOTE that 56 is an even number. The question never says even digits but just says even numbers.
  Hence num of ways to fill last place =4.
  As we have 7 digits in all and 0 cannot be in the initial position,
  Thus required number of ways = 4*5*5*4*3*2 = 100*24 = 2400 ways
• 11 years agoHelpfull: Yes(0) No(1)
• - - - - - -
  5*5*4*3*2*4=2400
  
• 11 years agoHelpfull: Yes(0) No(0)