Q. {1,2,3,4,5,5,6,6}how many four digited numbers can be formed without repiting the numbers?

• n*(n-1)*(n-2)...
  hence four digits with out repetition total 6 digits=n
  6*5*4*3=360
• 10 years agoHelpfull: Yes(21) No(5)
• 8*7*6*5/2*2=420
• 10 years agoHelpfull: Yes(11) No(2)
• we have to form 4 digit number
  _ _ _ _
  8*7*6*5
  as first place can hold any of the digits 1,2,3,4,5,5,6,6
  second place can hold 7 digits
  third place can hold 6 digits
  and fourth place can hold 5 digits
  since digit 5 is repeated 2 times and digit 6 is repeated 2 times
  
  required number of 4 digit number = (8*7*6*5)/2!*2! = 420
• 10 years agoHelpfull: Yes(10) No(0)
• Updated:
  
  1.nos containing only 1,2,3,4,5 and 6:6C4x4!=360
  2.nos containing two 5's(Ex:5512):(5C2x4!)/2!=120
  3.nos containing two 6's(Ex:6612):(5C2x4!)/2!=120
  4.nos containing two 5's and 6's(Ex:5566):(4!)/2!x2!=6
  Total nos:360+240+6=606
  
  
• 10 years agoHelpfull: Yes(8) No(1)
• 8P4/2!*2! AS THERE ARE 2 5S AND 2 6S
• 10 years agoHelpfull: Yes(7) No(3)
• ans=420
  arrangement of "n" object at "r" place where "r1" and "r2" are of common type
  =[n!/(n-r)!]*(1/r1!*r2!.....rn!)
  
  here
  n=8;
  r=4;
  r1=2;
  r2=2;
  ans=(8!/4!)(1/2!*2!);
  =(8*7*6*5)(1/4)
  =2*7*6*5
  =420
• 10 years agoHelpfull: Yes(4) No(0)
• guys just for a second imagine that there are only 1 5 and 1 6 so if we have 1,2,3,4,5,6 then for 4 digit no we can make following adjustment without any repetition
  for 1 st position we have 6 choice
  for 2 nd position we have 5 choice
  for 3 rd position we have 4 and for 4th we have 3....so no of possibility are 6*5*4*3...and now come back to asked question in which we have 2 5s and 2 6s...
  so final answer is 6*5*4*3/2!*2!
• 10 years agoHelpfull: Yes(2) No(0)
• nos containing only 1,2,3,4,5 and 6:6C4x4!=360
• 10 years agoHelpfull: Yes(1) No(1)
• total four digit no's are
  according to fundamental principle of counting
  four digit no= (8*7*6*5)/2!2!
  = 720 ans...
• 10 years agoHelpfull: Yes(0) No(6)
• 1.nos containing only 1,2,3,4,5 and 6:6C4x4!=360
  2.nos containing two 5's(Ex:5512):(5C2x4!)/2!=24
  3.nos containing two 6's(Ex:6612):(5C2x4!)/2!=24
  4.nos containing two 5's and 6's(Ex:5566):(4!)/2!x2!=6
  Total nos:360+48+6=414
  @sathvika:Is it right?
• 10 years agoHelpfull: Yes(0) No(4)
• 8c4/2!*2!
  Here for choosing 4 we have 8c4
  But for repetion divide with 2! for 5 and 6 value are repited
• 10 years agoHelpfull: Yes(0) No(4)
• are these Qs sufficent to crack the aptitude test of infosys?
• 10 years agoHelpfull: Yes(0) No(0)
• right answer for this question
  1,2,3,4,5,5,6,6
  4 different(1,2,3,4) 2 pair(5,5 6,6)
  I. both pair = 4!/2!2! = 6
  II. 1 pair and 2 different = 2c1*4c2*4!/2! = 288
  III. all different = 4! = 24
  total ways-
  = 24+288+6
  = 318 ans...
• 10 years agoHelpfull: Yes(0) No(0)