Exercise 6.2

Evaluate (i) 8 ! (ii) 4 ! - 3 !

Is 3 ! + 4 ! = 7 ! ? 3. Compute 8! 6! 2!× 4. If 1 1 6! 7! 8! x+ = , find x

Evaluate ( ) ! ! n n r− , when (i) n = 6, r = 2 (ii) n = 9, r = 5. 6.3.3 Derivation of the formula for nPr ( ) !P ! n r n n r−= , 0 ≤ r ≤ n Let us now go back to the stage where we had determined the following formula: nPr = n (n - 1) (n - 2) . . . ( n - r + 1) Multiplying numerator and denomirator by (n - r) (n - r - 1) . . . 3 × 2 × 1, we get ( ) ( ) ( )( )( ) ( )( ) / 1 2 1 1 3 2 1P 1 3 2 1 / n / r n n n ... n r n r n r ... n r n r . .. − − − + − − − × ×= − − − × × = ( ) ! ! n n r− , Thus ( ) !P ! n r n n r= − , where 0 < r ≤ n This is a much more convenient expression for nPr than the previous one. In particular, when r = n, !P ! 0! n n n n= = Counting permutations is merely counting the number of ways in which some or all objects at a time are rearranged. Arranging no object at all is the same as leaving behind all the objects and we know that there is only one way of doing so. Thus, we can have n P0 = 1 = ! ! ! ( 0)!= − n n n n ... (1) Therefore, the formula (1) is applicable for r = 0 also. Thus ( ) !P 0 ! n r n , r nn r= ≤ ≤− .