The number of permutations of the digits 1, 2 | vedclass

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(D) Total number of permutations of $7$ distinct digits is $7! = 5040$. Let $A$ be the set of permutations containing the string $153$. Treating $153$

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$4898$

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(D) Total number of permutations of $7$ distinct digits is $7! = 5040$. Let $A$ be the set of permutations containing the string $153$. Treating $153$ as a single block,we have $5$ items to arrange: ${153, 2, 4, 6, 7}$. Thus,$n(A) = 5! = 120$. Let $B$ be the set of permutations containing the string $2467$. Treating $2467$ as a single block,we have $4$ items to arrange: ${2467, 1, 3, 5}$. Thus,$n(B) = 4! = 24$. Let $A \cap B$ be the set of permutations containing both strings $153$ and $2467$. Treating these as two separate blocks,we have $2$ items to arrange: ${153, 2467}$. Thus,$n(A \cap B) = 2! = 2$. By the Principle of Inclusion-Exclusion,the number of permutations containing at least one of the strings is $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 120 + 24 - 2 = 142$. The number of permutations containing neither string is $Total - n(A \cup B) = 5040 - 142 = 4898$.

The number of permutations of the digits $1, 2, 3, ..., 7$ without repetition,which neither contain the string $153$ nor the string $2467$,is $........$.

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