Let v1 = 1, v2 = 4, v3 = 7, v4 = 6, v5 = 12. Let u1, u2, u3, u4, u5 represent the same numbers in a different order. Consider the product P = (u1 - v1)(u2 - v2)(u3 - v3)(u4 - v4)(u5 - v5), Then
option
a) P is always even.
b) P is always negative
c) P is always positive
d) P is even or odd depending on the ordering
e) P is always odd

• p is always even
• 11 years agoHelpfull: Yes(32) No(11)
• there are no fixed rule for this type of question
  we have to solve this question with the help of option
  a)take order of u1,u2,...=12,1,4,7,6 then p=(11)*(-3)*(-3)*(1)*(-6)=odd
  option a wrong
  
  b)take order of u1,u2,....=12,6,1,7,4 then=p=11*2*-6*1*-8= +ve
  option b wrong
  
  c)take order of u1,u2,...=12,1,4,7,6 then p=(11)*(-3)*(-3)*(1)*(-6)= -ve
  option c wrong
  
  d)take order of u1,u2,....=7,6,1,12,4 then p=6*2*-6*6*-8=even no
  correct ans because in option 1 we proved that it may be odd no ,and here we proved that it may be even number.
  
  e)in option d we proved that it may be even no
  wrong option
  
  
  final ans will be (d)
  
• 11 years agoHelpfull: Yes(24) No(10)
• ans is : pis always even .
• 11 years agoHelpfull: Yes(4) No(4)
• P is always even.
  P is always negative.
  
  both the above statements are true.
• 11 years agoHelpfull: Yes(3) No(15)
• Someone Please explain how P is always even
  Let we change the order as 4 1 7 6 12
  So P=3*(-3)*0*0*0=0 and 0 is odd....there are many cases where P is odd
• 11 years agoHelpfull: Yes(2) No(5)
• p is always even
  
• 11 years agoHelpfull: Yes(2) No(1)
• p is always even
• 11 years agoHelpfull: Yes(2) No(0)
• 0 is not an odd number @tanu
• 11 years agoHelpfull: Yes(1) No(1)
• @shikha gupta oh ya thnx for pointing it out bt still there are other cases where P can b odd.
• 11 years agoHelpfull: Yes(1) No(1)
• there are no fixed rule for this type of question
  we have to solve this question with the help of option
  a)take order of u1,u2,...=12,1,4,7,6 then p=(11)*(-3)*(-3)*(1)*(-6)=odd
  option a wrong
  
  b)take order of u1,u2,....=12,6,1,7,4 then=p=11*2*-6*1*-8= +ve
  option b wrong
  
  c)take order of u1,u2,...=12,1,4,7,6 then p=(11)*(-3)*(-3)*(1)*(-6)= -ve
  option c wrong
  
  d)take order of u1,u2,....=7,6,1,12,4 then p=6*2*-6*6*-8=even no
  correct ans because in option 1 we proved that it may be odd no ,and here we proved that it may be even number.
  
  e)in option d we proved that it may be even no
  wrong option
  
  
  final ans will be (d)
• 11 years agoHelpfull: Yes(1) No(0)