Think Logically not mathematically [part-33]

Algebra Level 3

How many ordered pairs ( x , y ) (x,y) of distinct non-negative integers exists such that x ! y ! = 1 \dfrac{x!}{y!}=1 ?

Notation: ! ! denotes the factorial function . For example: 8 ! = 1 × 2 × 3 × . . . × 8 8! = 1 \times 2 \times 3 \times ... \times 8 .

2 ∞ \infty 4 3 1 0

Sudoku Subbu by sudoku subbu , 19, India

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1 solution

Sudoku Subbu
Jun 26, 2016

We know that 1 ! = 0 ! = 1 1!=0!=1 a n d a l s o 1 ! 0 ! = 0 ! 1 ! = 1 1 = 1 and \space also \space \frac{1!}{0!}=\frac{0!}{1!}=\frac{1}{1}=1 Since X and Y are Distinct Whole numbers , The No.of Possible Values of ( X , Y ) (X,Y) is ( 0 , 1 ) (0,1) , 1 , 0 1,0 Therefore the number of possible values is 2.

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