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