Functions (XII)

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Question

$Let A = \{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\} and f : A \to A . The number of bijective functions such that$
$f (x_{i}) = x_{i} for exactly two of the x_{i}^{'} s is (i = 1 to 5)$

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Solution

$Number of ways of two elements follow f (x_{i}) = x_{i} is 5_{C_{2}}$
$Rest 3 unit does not satisfy f (x_{i}) = x_{i}$
$\begin{array}{l} = Number of de-arrangement of 3 things \\ = 3! (1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!}) = 3! (\frac{1}{2} - \frac{1}{6}) = 6 \times \frac{3 - 1}{6} = 2 \end{array}$
$Required no of ways = 10 \times 2 = 20$

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Functions (XII)

Remember concepts with our Masterclasses.

Let A=x1,x2,x3,x4,x5 and f:A→A . The number of bijective functions such that fxi=xi for exactly two of the xi′s is (i=1 to 5)

Number of ways of two elements follow fxi=xi is 5C2 Rest 3 unit does not satisfy fxi=xi= Number of de-arrangement of 3 things =3!1−11!+12!−13!=3!12−16=6×3−16=2 Required no of ways =10×2=20

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$Let A = \{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\} and f : A \to A . The number of bijective functions such that$
$f (x_{i}) = x_{i} for exactly two of the x_{i}^{'} s is (i = 1 to 5)$