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 (integer)
4898
Solution
Given that digits are $1,2,3,4,5,6,7$
<br/><br/>Total permutations $=7$!
<br/><br/>Let $p=$ Number which containing string 153
<br/><br/>$q=$ Number which containing string 2467
<br/><br/>$$
\begin{array}{ll}
& \therefore n(p)=5! \times 1 \\\\
& \Rightarrow n(q)=4! \times 1 \\\\
& \Rightarrow n(p \cap q)=2!
\end{array}
$$
<br/><br/>$$
\begin{aligned}
& \therefore n(p \cup q)=n(p)+n(q)-n(p \cap q) \\\\
& = 5 !+4 !-2 !=120+24-2=142
\end{aligned}
$$
<br/><br/>$\therefore n$ (neither string 143 nor string 2467)
<br/><br/>$=7 !-142=5040-142=4898$
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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