The number of permutations of the letters a b c d e f g such that neither the pattern beg

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
10 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?
9 years agoHelpfull: Yes(2) No(0)
4806
7!-2.5!+3!=4806
10 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)

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