A = {x_1, x_2, x_3, x_4}; B = {y_1, y_2, y_3, | vedclass

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(B) This is a problem of finding the number of derangements of $4$ elements.Let the set $A = \{x_1, x_2, x_3, x_4\}$ and $B = \{y_1, y_2, y_3, y_4\}$.

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) This is a problem of finding the number of derangements of $4$ elements.Let the set $A = \{x_1, x_2, x_3, x_4\}$ and $B = \{y_1, y_2, y_3, y_4\}$.$A$ one-one function $f: A 	o B$ is equivalent to a permutation of the set $\{y_1, y_2, y_3, y_4\}$.The condition $f(x_i) 
eq y_i$ for all $i \in \{1, 2, 3, 4\}$ means that no element is mapped to its corresponding index.This is the definition of a derangement of $n = 4$ objects,denoted as $D_n$.The formula for derangements is $D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots + (-1)^n \frac{1}{n!} ight)$.For $n = 4$:$D_4 = 4! \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} ight)$$D_4 = 24 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} ight)$$D_4 = 12 - 4 + 1 = 9$.Therefore,the number of such one-one functions is $9$.

$A = \{x_1, x_2, x_3, x_4\}$; $B = \{y_1, y_2, y_3, y_4\}$. $A$ function $f: A \to B$ is defined. The number of one-one functions such that $f(x_i) \neq y_i$ for $i = 1, 2, 3, 4$ is equal to:

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