Latest Elitmus Aptitude Question SOLUTION: 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

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(#M40039277) ELITMUS Numerical Ability question permutation and combination Keep an EYE

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

Asked In Elitmus amit kumar (9 years ago)

Unsolved Read Solution (6)

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Read Solution (6) :

Devang  (9 years ago)

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

Is this solution Helpfull? Yes (56) | No (2)  

veer kumar  (9 years ago)

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

Is this solution Helpfull? Yes (9) | No   

Aravindkumar  (9 years ago)

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.

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Anamika  (8 years ago)

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?

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durga devarapu  (9 years ago)

4806
7!-2.5!+3!=4806

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Amit Kumar Singh  (7 years ago)

I think the answer would be
7!- 3!*3!*3!= 4824
but not given in the option ..

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