if a die has 1 6 and 3 4 and 2 5 opposite each other how many such dies can be made

• ans 12
  if one face have 1,3,2, then the crossponding face will be occupied by respective pair i.e (6,4,5).
  so there r 3 place and we have 3 place soit can be arranged in 3! way i.e 6 way and every pair will be suffled in 2 way so ans will be6*2=12.
• 8 years agoHelpfull: Yes(23) No(2)
• Ans::24 ways
  
  if the dice shows 6 on top then 1 will be at the bottom.
  
  so the remaining 3,4 and 2,5 can be plotted in 4 ways so four different dies possible
  
  if 1 is placed on the top then again 4 ways
  
  
  4*2+4*2+4*2=24 ways
• 9 years agoHelpfull: Yes(4) No(6)
• 12
  there are 3 pairs : (1,6),(3,4),(2,5)
  and each pair has 2 sides
  so, 3*2*1*2 = 12
• 9 years agoHelpfull: Yes(4) No(4)
• 
  2!+2!+2!=6
  all can be arranged in 3! ways
  hence 36
  
  
• 8 years agoHelpfull: Yes(2) No(1)
• 3!*2*2*2=48
• 8 years agoHelpfull: Yes(2) No(0)
• whats the correct ans
• 8 years agoHelpfull: Yes(2) No(0)
• permutation and combination
• 8 years agoHelpfull: Yes(1) No(1)
• Ans: 8
  Taking 1 at one of its faces, the other possibilities for other numbers 2,3,4,5 is 4.
  Now interchanging 1 and 6 (2 ways), we will have 4 more possibilities.
  
  Hence, total number if ways: 4*2=8
• 8 years agoHelpfull: Yes(1) No(2)
• Let's take a blank die which has faces a,b,c,d,e,f
  
  (a,b),(c,d),(e,f) are opposite pair of sides
  
  (1,6),(3,4),(2,5) are 3 possible opposite side sets required
  
  These three can be arranged in above blank cube in 3*2 = 6 ways
  
  Each of these pairs can be arranged in 2 ways (1,6) or (6,1).
  
  Therefore Total number of dice possible = 6*2 = 12
• 6 years agoHelpfull: Yes(1) No(0)
• if take 1 and 6 as fixed then total 8 posiblities of (3,4) and (2,5)
  then alter 1 and 6 then again 8 posoblities
  same for (3,4) and (2,5)
  total 48 posiblities but here some 8 posoblities is repeated so answer is:40
• 8 years agoHelpfull: Yes(0) No(1)
• Let us assume their r 2 configurations at a vertice
  A. and. A
  B. C. C. B
  
  
  If we interchange A and B in the second configuration then it becomes the first configuration
  
  This means two possible basic arrangements of the the 3 sets at a vertice
  
  Now, 1_6 ,3_4, 5_2. Can interchange.
  
  
  This gives total ways as 2*2*2*2
• 8 years agoHelpfull: Yes(0) No(1)
• 8 ways = 2*2*2
  1 6 1 6 1 6 1 6
  3 4 3 4 4 3 4 3
  2 5 5 2 2 5 5 2
  
  6 1 6 1 6 1 6 1
  3 4 4 3 4 3 3 4
  2 5 2 5 5 2 5 2
• 6 years agoHelpfull: Yes(0) No(0)