Let $A = \left\{ x_{1}{,x}_{2,}x_{3},\ x_{4},\ x_{5},\ x_{6} ight\}$, $B = \left\{ y_{1},\ y_{2},\ y_{3},y_{4},y_{5},y_{6} ight\}.$Then the number of one –one mapping from $A$ to $B$ such that $f\left( x_{i} ight)
eq y_{v}i = 1,\ 2,\ 3,\ 4,\ 5,\ 6$ is

Question

Let $A = \left\{ x_{1}{,x}_{2,}x_{3},\ x_{4},\ x_{5},\ x_{6} ight\}$, $B = \left\{ y_{1},\ y_{2},\ y_{3},y_{4},y_{5},y_{6} ight\}.$Then the number of one –one mapping from $A$ to $B$ such that $f\left( x_{i} ight) 
eq y_{v}i = 1,\ 2,\ 3,\ 4,\ 5,\ 6$ is

Dr. Abubacker · Accepted Answer

(b)
$\because f(x_{i}) 
eq y_{i}$
$ie,$ no object goes to its scheduled place. Then, number of one-one mappings
$= 6!\left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{6!} ight)$
$= 6!\left( \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{6!} ight)$
$= 360 - 120 + 30 - 6 + 1 = 265$

Q. Let $A = \left\{ x_{1}{,x}_{2,}x_{3},\ x_{4},\ x_{5},\ x_{6} \right\}$, $B = \left\{ y_{1},\ y_{2},\ y_{3},y_{4},y_{5},y_{6} \right\}.$Then the number of one –one mapping from $A$ to $B$ such that $f\left( x_{i} \right) \neq y_{v}i = 1,\ 2,\ 3,\ 4,\ 5,\ 6$ is

Explanation

Reviewed by Dr. Abubacker

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