The number of permutations of the letters a,b,c,d,e,f,g such that neither the pattern 'beg' nor 'acd' occurs is
a.4806
b.420
c.2408
d.non of these

• Here,
  Ans
  = (Total Ways)-(No. Of Ways in which 'beg' occurs)-(No. of ways in 'acd' occurs) + (No. of ways in which both 'beg'&'acd' occurs)
  
  = (7! total ways) - (5! 'beg' occurs) - (5! 'acd' occurs) + (3! both occurs)
  
  = (5040) - (120) -(120) + (6)
  
  = 5040 - 240 + 6
  
  = 4806-------> ANS
• 9 years agoHelpfull: Yes(56) No(2)
• 1.total arrangement=7!=5040
  2.beg occurs but not acd=5!-3!=114
  3.acd occurs but not beg=5!-3!=114
  4.beg and acd both occurs=3!=6
  answer will=(1)-(2+3+4)=(5040)-(114+114+6)=4806
• 9 years agoHelpfull: Yes(9) No(0)
• We will subtract the restricted cases from the total number of cases.
  Total number of cases = 7! = 5040
  Cases when ‘beg’ occurs: (beg)_ _ _ _ = 5! = 120
  Cases when ‘acd’ occurs: (acd’)_ _ _ _ = 5! = 120
  But there would be cases in the above permutations where both ‘beg’ and ‘acd’ occur together. The number of such cases = 3! = 6
  So, the final number of cases = 7! – 5! – 5! + 3! = 5040 – 120 – 120 + 6 = 4806.
• 9 years agoHelpfull: Yes(6) No(0)
• but why we will add "no. of ways in which both beg and acd occurs" rather we should subtract it na bcause the question says it should neither have any of the patterns.. so we are adding up 3! ?
  Please explain anyone?
• 8 years agoHelpfull: Yes(2) No(0)
• 4806
  7!-2.5!+3!=4806
• 9 years agoHelpfull: Yes(1) No(0)
• I think the answer would be
  7!- 3!*3!*3!= 4824
  but not given in the option ..
• 7 years agoHelpfull: Yes(0) No(0)