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Let a function f : XtoY is defined where X={0,1,2,3,….,9}, Y={0,1,2,…..,100} and f(5)=5, then the probability that the function of type f: xtoB where BsubeY is of bijective in nature is |
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Answer» `(10!)/(sum_(r=1)^(101)r^(9)*^(100)C_(r-1))` For total cases `'B'` is set of one ELEMENT, no. of function `-1` `'B'` is set of two element, no. of function `-2^(9)` `'B'` is set of three element, no. of function `-3^(9)` ......... ........ `'B'= Y`, No. of function `=101^(9)` `:.` Total cases `=1+"^(100)C_(1)*2^(9)+^(100)C_(2)*3^(9)+....+^(100)C_(100)*101^(9)` `=sum_(r=1)^(101)r^(9)*"^(100)C_(r-1)` `:.` Required probability `=("^(100)C_(9)*9!)/(sum_(r=1)^(101)r^(9)*^(100)C_(r =1))` |
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