how many even natural number divisible by 5 can be formed using 0,1,2,3,4,5,6 with no repeatation of numbers

• 6p1+6p2+6p3+6p4+6p5+6p6
  =1956 Ans.
  
• 10 years agoHelpfull: Yes(25) No(6)
• Hi! the correct answer is 1956.
  Even number and divisible by five means: number has to end with 0.
  Here we need to consider all the case. they are elucidated as follows :
  1 digit number: none. ( as set of natural numbers doesn't include o)
  2 digit number:place 0 at the unit's place, and we are left with only 6 options for the ten's digit. so 6.
  3 digit numbers:--0. 6 options for hundred's place and 5 options for ten's place as repetitions are strictly forbidden.. so numbers possible :6*5=30. Similarly for the 4 digit , 5 digit, 6 digit and 7 digit numbers.
  4 digit number:---0. 6*5*4=120.
  5 digit numbers:----0. 6*5*4*3=360.
  6 digit numbers:-----0. 6*5*4*3*2=720
  7 digit numbers:------0. 6*5*4*3*2*1=720.
  No other possibilities remain as repetitions are not allowed. Hope this has clarified your concern.
  Regards.
• 10 years agoHelpfull: Yes(16) No(0)
• 1 digit no. =1(0)
  2 digit no. =_0 =6
  3 digit no. =_ _ 0 =6*5=30
  4 digit no. =_ _ _ 0=6*5*4=120
  5 digit no. =_ _ _ _ 0=6*5*4*3=360
  6 digit no. =_ _ _ _ _ 0=6*5*4*3*2=720
  7 digit no=_ _ _ _ _ _0=6*5*4*3*2*1=720
  total=1957
• 10 years agoHelpfull: Yes(15) No(6)
• 6p1+6p2+6p3+6p4+6p5+6p6=1956
  bcoz 2digit 3digit 4digit 5digit 6 digit and 7 digit numbers are considered
• 10 years agoHelpfull: Yes(3) No(0)
• plz notice that it has asked for even natural nos. so we are not to consider nos ending with 5. we have to form nos ending with 0 only. so fix 0 at the unit's place and then the rest 6 digits in 6! ways. so its 720
• 10 years agoHelpfull: Yes(3) No(2)
• bhai plz explain kaise hua mera 1236 a raha hai
• 10 years agoHelpfull: Yes(2) No(0)
• only no.s having last digit as 0 are both divisible by 5 and are even.
  so,
  case 1 : there are 0 no.s of 1 digits which is divisible by 5 and are even.
  case 2 : there are 6 no.s of 2 digits (as tens place can be fill by 6 no.s only)
  case 3 : there are 30 no.s of 3 digits (at hundreds place 6 no,tens place 5 no)
  case 4 : there are 120 no.s of 4 digits (thous place=6 , h = 5 , t = 4 )
  case 5 : there are 360 no.s of 5 digits (at ten thous place=6 ,t=5,h=4,t=3)
  similarly,
  case 6 : there are 720 no.s of 6 digits
  case 7 : there are 720 no.s of 7 digits
  hence total no.s are 0+6+30+120+360+720+1440 = 2676
• 10 years agoHelpfull: Yes(2) No(3)
• answer will be 720
• 10 years agoHelpfull: Yes(1) No(7)
• is it 720?
• 10 years agoHelpfull: Yes(1) No(5)
• 6!=720 EASY HUH!!!!!
• 10 years agoHelpfull: Yes(1) No(7)
• The no.s will be divisible by 5 only if last digit 0 or 5.
  As no number can begin with 0 so, 6!/2!=360
• 10 years agoHelpfull: Yes(1) No(3)
• 1 digit no. =1 (i.e 0)
  2 digit no. =_ 0 =5
  3 digit no. =_ _ 0 =5*4=20
  4 digit no. =_ _ _ 0=60
  5 digit no. =_ _ _ _ 0= 120
  6 digit no. = _ _ _ _ _ 0=120
  total= 326
• 10 years agoHelpfull: Yes(1) No(3)
• here 0 will be fixed at ones place
  so 6! = 720
• 10 years agoHelpfull: Yes(1) No(1)
• aditya kumar you added an extra 720.
  
  
• 10 years agoHelpfull: Yes(1) No(1)
• sorry i have by mistake added an extra 720 to my solution .....
  hence the correct answer is 1956.
• 10 years agoHelpfull: Yes(1) No(0)
• 6!/2! is the answer
• 10 years agoHelpfull: Yes(0) No(0)
• the last place is filled with 0. then the remaining 6 places can be filled in 6! ways= 720
• 10 years agoHelpfull: Yes(0) No(1)
• 6c1+6c2+6c3+6c4+6c5+6c6=2^6-1= 63 (no repetation of number)
• 10 years agoHelpfull: Yes(0) No(0)
• 7P2+7P6+7P5+7P4+7P3+7P2+7P1
• 10 years agoHelpfull: Yes(0) No(0)
• 1236
  6+6*5+6*5*4+6*5*4*3+6*5*4*3*2
• 10 years agoHelpfull: Yes(0) No(0)
• 6*5*4*3*2*1*1=720
  even natural no. divisible by 5 must end with 0 i.e at unit place it will contain 0 value.
• 10 years agoHelpfull: Yes(0) No(0)
• i m confused plzzzz...suggest which is correct answer'
  
• 10 years agoHelpfull: Yes(0) No(0)
• urvashi@ is quetion me ye nai kahe raha ki only 7 seven digit no it may be 2 digit , 3 digit n 4 digit upto 7 digit possiable ho sakta hai so Saurabh agarwal is right
  
• 10 years agoHelpfull: Yes(0) No(0)
• Don't get confused between the formation of 7 digit numbers or even natural numbers. correct answer is 1956 for even numbers comprising of 2 digit no...upto 7 digit.
  And 7 digit number divisible by 5 and even answer is 720
• 10 years agoHelpfull: Yes(0) No(0)
• case1: 1 digit=5=1 way
  case2: 2 digit=_ 5;_ 0=>(5*1)+(6*1)=11 because the numbers ending with 5 or 0 are divisible by 5
  case3: 3 digit=_ _ 0;_ _ 5=>(6*5*1)+(5*5*1)=55
  case4: 4 digit=_ _ _ 0;_ _ _ 5=>(6*5*4*1)+(5*5*4*1)=220
  case5: 5 digit=_ _ _ _ 0;_ _ _ _ 5=>(6*5*4*3*1)+(5*5*4*3*1)=660
  case6: g digit=_ _ _ _ _ 0;_ _ _ _ _ 5=>(6*5*4*3*2*1)+(5*5*4*3*2*1)=1320
  by adding all=>1320+660+220+55+11+1=2267.
  i think this is the process... correct me if am wrong
• 10 years agoHelpfull: Yes(0) No(0)