Question

0!=? explain in brief math expert question​

Answer 1

Consider the following expressions :

n × ( n - 1 ) × ( n - 2 ) × ... × { n - ( r - 1 ) }

= n × ( n - 1 ) × ( n - 2 ) × ... × ( n - r + 1 ) = P ( n , r )

= [ n × ( n - 1 ) × ( n - 2 ) × ... × ( n - r + 1 ) ] × (n-r ) ! / ( n-r ) !

Note : [ n × ( n - 1 ) × ( n - 2 ) × ... × ( n - r + 1 ) ] × (n-r ) ! =

[ n × ( n - 1 ) × ( n - 2 ) × ... × ( n - r + 1 ) ] × (n-r ) × ( n - r -1 ) × ( n - r - 2 ) ×... × 3 × 2 × 1 = n !

Thus we must have

[ n × ( n - 1 ) × ( n - 2 ) × ... × ( n - r + 1 ) ] × (n-r ) ! / ( n-r ) ! = n ! / ( n - r ) !

The expression n ! / ( n - r ) ! can be written as

P ( n , r ) = n ! / ( n - r ) !

If n = r ,

P ( n , r ) = P ( n , n ) = n ! / ( n - r ) ! = n ! / 0 !

=> P ( n , n ) = n ! / 0!

=> 0! = n ! / P ( n , n ) ----- ( i )

Now P ( n , r ) = n × ( n - 1 ) × ( n - 2 ) × ... × ( n - r + 1 )

if n = r then ,

P ( n , n ) = n × ( n - 1 ) × ( n - 2 ) × ... × ( r - r + 1 )

= n × ( n - 1 ) × ( n - 2 ) × ... × ( 1 )

= n × ( n - 1 ) × ( n - 2 ) × ... × 3 × 2 × 1

= n !

=> P ( n , n ) = n ! ------ ( ii )

Now from equation ( i ) we have ,

0 ! = n ! / P ( n , n )

= n ! / n ! ------ using equn. ( ii )

= 1

Hence 0 ! = 1