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The number of permutations of letters a,b,c,d,e,f,g so that ne

Question

The number of permutations of letters $a, b, c, d, e, f, g$ so that neither the pattern $b e g$ nor $c a d$ appears is

\frac{7 !}{3 ! 3 !}

\frac{7 !}{2 ! 3 ! 3 !}

4806

None of these

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Elijah Discussed

The number of permutations of letters

a, b, c, d, e, f, g

so that neither the pattern

b e g

nor

c a d

appears is

6 mins ago

Discuss this question LIVE

6 mins ago

Text solutionVerified

The total number of permutations

= 7!

Let A be the property that

b e g

occurs and B be the property that

c a d

occurs.

Number of permutations with A

= 5!

=that of with B and

n (A \cap B)

= 3!

Thus

n (A \cap B) = 5! + 5! - 3! = 234

Thus required number

= 7! - 234 = 4806

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Question Text	The number of permutations of letters $a, b, c, d, e, f, g$ so that neither the pattern $b e g$ nor $c a d$ appears is
Answer Type	Text solution:1
Upvotes	150