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प्रश्न
One mapping is selected at random from all mappings of the set A = {1, 2, 3, ..., n} into itself. The probability that the mapping selected is one to one is
पर्याय
\[\frac{1}{n^n}\]
\[\frac{1}{n!}\]
\[\frac{\left( n - 1 \right)!}{n^{n - 1}}\]
None of these
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उत्तर
Number of ways to map 1st element in set A = n
Number of ways to map 2nd element in set A = n and so on
∴ Total number of mapping from set A to itself = \[n \times n \times . . . \times n\] (n times) = \[n^n\]
For one to one mapping,
Number of ways to map 1st element in set A = n
Number of ways to map 2nd element in set A = n −1
Number of ways to map 3rd element in set A = n − 2
. . . . . . . .
. . . . . . . .
Number of ways to map nth element in set A = 1
Total number of one to one mappings from set A to itself = \[n \times \left( n - 1 \right) \times \left( n - 2 \right) \times . . . \times 1 = n!\]
∴ Required probability = \[= \frac{\text{ Total number of one to one mappings from set A to it self } }{\text{ Total number of mappings from set A to it self} } = \frac{n!}{n^n} = \frac{\left( n - 1 \right)!}{n^{n - 1}}\]
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