how many 4 digit no. divisible by 5 can be formed with the digits 0,1,2,3,4,5,6 and 6..
option
1) 432
2) 249
3) 220
4) 288

• it cannot be other than 249
  0, 1, 2, 3, 4, 5, 6, 6
  case 1 : __ __ __ 0
  all are different(1,2,3,4,5,6) = 6C3 * 3! = 120
  1 pair of 6 and 1 other number from (1,2,3,4,5) = 5C1 * 3!/2 = 15
  
  case 2 : __ __ __ 5
  1 pair of 6 and 1 other number from (1,2,3,4)(not 0) = 4C1 * 3!/2 = 12
  1 pair of 6 and 1 zero(0) = 2 (0 cannot be at the first place, u need 4 digit number)
  all are different(0,1,2,3,4,6) = 1 place(1,2,3,4,6); 2,3 place(0,1,2,3,4,6) = 5*5*4 =100
  
  = 120+15+12+2+100 = 249
• 10 years agoHelpfull: Yes(22) No(2)
• 588 with repetition(6*7*7*2)
  220 without repetition((6*5*4*1)+(5*5*4*1))
• 10 years agoHelpfull: Yes(20) No(8)
• ans is 220
• 10 years agoHelpfull: Yes(16) No(5)
• The answer is 249
  first of all place the non repeated digits ending with 0 so we get 6p3=120
  by placing last with 5 we get 1000 place can be fill with 5 possibles and remaing with 5p2 so we get =5*5p2=100
  
  
  upto know we get values having without reputation of 6
  Now we can go with reputation of 6 ........
  first place the unit digit with the 0,we place 2 sixes and one is different and this can be arranged in slecting the different of remaing is 5c1*3!/2!==15
  as because six is repeated 2 times then we are going to divide with the 2!
  
  
  by placing 5 at last we have two posibilities
  first place 6 at first then another 6 in other , then other value from 5c1*2!=10
  here 2! because 6 and different number can be arranged in 2! ways...
  last thing not starting with 6 by placing the last digit value 5 is
  
  first place can we fill in 4c1 ways and remaing two with 6 so the posibilities are 4c1*1=4
  
  
  By adding all the total we get 249.
  
• 10 years agoHelpfull: Yes(11) No(2)
• Without repetition of 6:
  Ending with 5 is 100 nos
  Ending with 0 is 120 nos
  
  With repetition of 6:
  Ending with 5 is 14 nos
  Ending with 0 is 15 nos
  
  Total nos are: 249
• 10 years agoHelpfull: Yes(8) No(2)
• answer is 432..
  reason:- Since in order to get divisible by 5 ,the units place must be vth 0 or 5 so it is 2! and d remaining nos r 6 and in tens plac dey can b arranged in 6! and hundreds plc 5! and thousands plc 4!=4!+5!+6!+2!=865/2=432.. since 6 is twice.
• 10 years agoHelpfull: Yes(7) No(5)
• ans is 249
• 10 years agoHelpfull: Yes(5) No(1)
• 6*7*7*2 = 588 Ans.
• 10 years agoHelpfull: Yes(1) No(8)
• koi shi solution btao iska??
• 10 years agoHelpfull: Yes(1) No(2)
• Ans;3) 220
  Sol ; Case I, when the last digit is fix 0 (zero)
  5*5*4*1=100
  Case II,when the last digit is fix 5
  6*5*4*1=120
  total 4 digit no divisible by 5 is
  Case I+Case II=100+120=220
• 8 years agoHelpfull: Yes(1) No(3)
• I think answer should be 896
• 10 years agoHelpfull: Yes(0) No(7)
• 432.
  On unit place we can put (0,5).So we can place on unit place by 2 ways,and other three places we can submit by 6 ways,because repeatation is alowed.
  Ans: 6*6*6*2=432
• 10 years agoHelpfull: Yes(0) No(6)
• arun sharma's permutation and combination lod 2 2nd question
• 7 years agoHelpfull: Yes(0) No(0)
• Unit digit can be chose in 2 ways- either 0 or 5.
  
  When unit digit is 5:
  With just a single 6:
  
  Number of numbers possible = 5 * 5 * 4 = 100 (as MSD cannot be a 0)
  
  Now, two 6's can come as 66x 6x6 or x66, and this counts to 5 + 5 + 4 = 14 numbers (for the last case, x can only be from 1-4 and not 0)
  
  So, total numbers ending in 5 = 100 + 14 = 114.
  
  When unit digit is 0:
  With just a single 6:
  
  Number of numbers possible = 6 * 5 * 4 = 120
  
  Now, two 6's can come as 66x 6x6 or x66, and this counts to 5 + 5 + 5 = 15 numbers.
  
  So, total numbers ending in 0 = 120 + 15 = 135.
  
  Thus, total numbers divisible by 6 = 114 + 135 = 249.
  
• 7 years agoHelpfull: Yes(0) No(0)
• The remainder we get when we divide the sum of the digits by 9 is the same as the remainder when we divide the number by 9
  so rem of 22*22/9 = rem of 4*4/9
  rem of 222*222/9 = rem of 6*6/9
  
  rem of 2^2+ 4^2+6^2.............98^2/9
  rem of 2^2(1+2^2+3^2............+49^2)/9
  
  1+2^2+3^2............+49^2=49(49+1)(98+1)/6=40,425
  
  rem of 4*40425/9 = 161700/9= 1+6+1+7=15
  Rem=1+5=6
• 6 years agoHelpfull: Yes(0) No(0)