Permutations with Repetitions Examples on Numbers2

In this class, We discuss Permutations with Repetitions Examples on Numbers2.

The reader should have prior knowledge of permutations with repetitions formulae. Click Here.

Example:

The given digits are 2, 3, 6, 6, and 7.

1) Find the count of numbers less than 60000.

To find the numbers less than 60000, we need to place 2 or 3 in the first position.

The first position is filled in two ways.

The remaining four positions are filled in 4!/2! ways.

Total permutations are 2 * 4!/2!.

2) Total numbers less than 70000.

The first position is filled with 2, 3, or 6.

Element 6 is repetition. So, find permutations separately.

By placing 2 or 3 in the first position, we have 2 * 4!/2! permutations.

By placing 6 in the first position, we have 4! permutations.

Total = 2 * 4!/2! + 4!.

3) Numbers divisible by 4.

if the last two digits are divisible by four, then the number is divisible by four.

We have the following options in the last two digits.

32, 36, 72, and 76 are the four possibilities to place in the last two digits.

The above four numbers are divisible by four.

For 32, the permutations are 3!/2!

For 36, the permutations are 3!

For 72, the permutations are 3!/2!

For 76, the permutations are 3!

Total permutations = 3!/2! + 3! + 3!/2! + 3!.