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(B) We need to find the number of permutations of ${1, 2, 3, 4}$ such that the patterns $12, 23, 34$ do not occur.Total permutations of $4$ digits is

Vedclass · Accepted Answer

$11$

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(B) We need to find the number of permutations of ${1, 2, 3, 4}$ such that the patterns $12, 23, 34$ do not occur.Total permutations of $4$ digits is $4! = 24$.Let $S$ be the set of all permutations. Let $A_1$ be the set of permutations containing $12$,$A_2$ be the set containing $23$,and $A_3$ be the set containing $34$.We want to find $|S| - |A_1 \cup A_2 \cup A_3|$.Using the Principle of Inclusion-Exclusion:$|A_1| = |A_2| = |A_3| = 3! = 6$.$|A_1 \cap A_2|$ (contains $123$) = $2! = 2$.$|A_2 \cap A_3|$ (contains $234$) = $2! = 2$.$|A_1 \cap A_3|$ (contains $12$ and $34$) = $2! = 2$.$|A_1 \cap A_2 \cap A_3|$ (contains $1234$) = $1! = 1$.$|A_1 \cup A_2 \cup A_3| = (6+6+6) - (2+2+2) + 1 = 18 - 6 + 1 = 13$.Number of valid permutations = $24 - 13 = 11$.

The total number of $4$-digit numbers which can be formed using the digits $1, 2, 3, 4$ without repetition such that the digit $n+1$ never immediately follows the digit $n$ (for $n=1, 2, 3$) is:

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