JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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

Choose the correct answer:

Explanation

Given numbers are 1, 2, 3, 4, 5, 6, 7
$So, total number of permutations are 7! = 5040$
Now total number of permutations having string (153)
$= (153), 2, 4, 6, 7$
$\Rightarrow n (153) = 5! = 120$
$and total number of permutations having string (2467)$
$= (2467), 1, 3, 5$
$\Rightarrow n (2467) = 4! = 24$
$also total permutations having string (153) and (2467)$
$n (153 \cap 2467) = 2! = 2$
$Now n (153 \cup 2467) = 120 + 24 - 2 = 142$
$So, the required no. of permutations are$
$n (\overline{153} \cap \overline{2467}) = n (\overline{153 \cup 2467})$
$= Total - n (153 \cup 2467)$
$= 5040 - 142 = 4898$

Choose the correct answer:

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