permutations with repetition and restrictions

Today in the class we started Permutations with repetitions and restrictions. To do these questions we have to remember Permutation Part1 and Permutations Part 2.
We have to remember:
• When there are no restrictions then we can use either dash method or formula.
Formula
nPr = n! / (n-r)!

• If there are restrictions then we have to use dash method only.
• If there are any restriction then we have to use formula: n!/a!b!c!d! or we could use dash method and divide it by a!b!c!d!.

For example:
A family of 7 consisting of 2 parents and 5 children are going to be arranged for a photo:

a) If there are no restrictions on how they sit, how many possible arrangements are there?
We can solve this question by dash met
7 . 6 . 5 .4 . 3 . 2 . 1 = 5040
This can also be written as 7! That will also give us same answer.

b) IF parents must be seated together, how many possibilities are there?
For parents: There are two parents so two dashes and we also have two seats for them. So
2 . 1 = 2!but parents should be together so they can be called “1 Unit”

For the kids: There are 5 kids so 5 ashes and we have 5 seats or them. So
5 . 4 . 3 . 2 . 1 =5 !but kids can sit anywhere they want so they can be represented as “5 Units”

So now total Units =6!
Possibilities: 6! . 2! = 1440

c) If parents cannot be seated together, how many possibilities are there?
Total possibilities – If parents sit together
5040 - 1440
= 3600

d) If parents and oldest child must be together, how many arrangements are possible?
e) For parents and oldest child: There are 2 parents and a child so three dashes and 3 seats for them .
3 . 2 . 1 =3!these can be represented as “1Unit”

For kids: if one child should sit with parents so there are 4 left left, then there should be 4 dashes and 4 seats for them.
4 . 3 . 2 . 1 = 4!
Total Units =5!
Possibilities
5! . 3! = 720