First Chair can be occupied by any of 6 persons , that is 6 choices
Second Chair can be occupied by any of 5 remaining persons , that is 5 choices
Third Chair can be occupied by any of 4 remaining persons , that is 4 choices
So , number of possible ways is 6 * 5 * 4 = 120
Now let ’ s solve this using the formula . Number of possible arrangement is n Pr
In this case n = 6 & r = 3 . Applying this in formula we get Number of arrangement is 6 P3 , that is ( 6 !)/( 6-3 )! = 120
Theorem 2 : Number of permutations of n different objects taken r at a time , where repetition is allowed , is n r .
Numerical : Find 3 digit numbers formed from ( 1,3,5,7,9 ) when repetition is allowed