Let a function \( f: X \rightarrow Y \) is defined where \( X=\{0,1,2,3, \ldots, 9\} \), \( Y=\{... VIDEO
Let a function \( f: X \rightarrow Y \) is defined where \( X=\{0,1,2,3, \ldots, 9\} \), \( Y=\{0,1,2, \ldots, 100\} \) and \( f(5)=5 \), then the probability that the function of type \( f: x \rightarrow B \) where \( B \subseteq Y \) is of bijective in nature is
(a) \( \frac{10 \text { ! }}{\sum_{r=1}^{101} r^{9} \cdot{ }^{100} C_{r-1}} \)
(b) \( \frac{{ }^{101} C_{9} \cdot 9 \text { ! }}{\sum_{r=1}^{101} r^{10} \cdot{ }^{100} C_{r}} \)
(c) \( \frac{{ }^{100} C_{9} \cdot 9 \text { ! }}{\sum_{r=1}^{101} r^{10} \cdot{ }^{101} C_{r}} \)
(d) \( \frac{{ }^{100} C_{9} \cdot 9 !}{\sum_{r=1}^{101} r^{9} \cdot{ }^{100} C_{r-1}} \)
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