The number of permutations, of the digits 1, 2, 3, ..., 7 without repetition, which neither contain the string 153 nor the string 2467, is ________ . [2023]

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Solution

Q.
The number of permutations, of the digits 1, 2, 3, ..., 7 without repetition, which neither contain the string 153 nor the string 2467, is ________ . [2023]

Ans.
(4898)

Given digits are 1, 2, 3, 4, 5, 6, 7

Total number of permutations = 7!

Let A = number of numbers containing string 153

Let B = number of numbers containing string 2467

$n (A) = 5! \times 1, n (B) = 4! \times 1, n (A \cap B) = 2!$

Required number of permutations = Total number of permutations - $n (A \cup B)$

$= 7! - (5! + 4! - 2!) = 7! - 142 = 4898$

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