JEE MAIN - Mathematics (2023 - 10th April Morning Shift - No. 20)
The number of permutations, of the digits 1, 2, 3, ..., 7 without repetition, which neither contain the string 153 nor the string 2467, is ___________.
Answer
4898
Explanation
Given that digits are $1,2,3,4,5,6,7$
Total permutations $=7$!
Let $p=$ Number which containing string 153
$q=$ Number which containing string 2467
$$ \begin{array}{ll} & \therefore n(p)=5! \times 1 \\\\ & \Rightarrow n(q)=4! \times 1 \\\\ & \Rightarrow n(p \cap q)=2! \end{array} $$
$$ \begin{aligned} & \therefore n(p \cup q)=n(p)+n(q)-n(p \cap q) \\\\ & = 5 !+4 !-2 !=120+24-2=142 \end{aligned} $$
$\therefore n$ (neither string 143 nor string 2467)
$$ =7 !-142=5040-142=4898 $$
Total permutations $=7$!
Let $p=$ Number which containing string 153
$q=$ Number which containing string 2467
$$ \begin{array}{ll} & \therefore n(p)=5! \times 1 \\\\ & \Rightarrow n(q)=4! \times 1 \\\\ & \Rightarrow n(p \cap q)=2! \end{array} $$
$$ \begin{aligned} & \therefore n(p \cup q)=n(p)+n(q)-n(p \cap q) \\\\ & = 5 !+4 !-2 !=120+24-2=142 \end{aligned} $$
$\therefore n$ (neither string 143 nor string 2467)
$$ =7 !-142=5040-142=4898 $$
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