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 nature is

Options

A. `(10!)/(sum_(r=1)^(101)r^(9)*^(100)C_(r-1))`
B. `("^(101)C_(9)*9!)/(sum_(r=1)^(101)r^(10)*^(100)C_(r))`
C. `("^(100)C_(9)*9!)/(sum_(r=1)^(101)r^(10)*^(101)C_(r))`
D. `("^(100)C_(9)*9!)/(sum_(r=1)^(101)r^(9)*^(100)C_(r-1))`

Answer

D

Explanation

`(d)` Favourable cases for bijection is `"^(100)C_(9)xx9!`
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))`