Permutations with Repetitions and Restrictions

Permutations: 

with Repetitions and Restrictions, and Case Restrictions

Hey class! I'm Ann and I'm going to talk about the things we covered today. HOW FUN!!!!!!

Anyways....

We covered two topics today, Permutations with Repetitions and Restrictions, and Permutations with Case Restrictions. 

Lets go back to our precious lessons. As we all know, permutation  is a set of distinct objects in an arrangement of objects, without repetition into a specific order. To permute a set of objects means to rearrange them.

We can use two methods to solve permutations problems:

• the permutation formula 
• Dash Method

Repetition means you are allowed to pick the same item more than once

Restrictions are when you are asked to place an item in a specific location

check out Carmelo and Billy`s blog post for more infos: http://precalculus40ssectioncwinter2015.blogspot.ca/2015/02/permutations-part-2-time-to-have-some.html 

Here are some EXAMPLES:

Permutations with Repetitions and Restrictions:

1)      Cathlene, JJ, PK, Aldrin, and Christy are going to be arranged for the Santa picture:

       a)If there is no restrictions on where they stand, how many arrangements are there

   n=5    r=5         

   5!= 120

       or

Dash method:

5• 4• 3• 2•1= 120

      b) If Christy (C) and PK (P) must be standing together, how many possibilities are there

2•1•3•2•1= 12 OR

CP

            3•2•1•2•1= 12 OR

                CP

            3•2•2•1•1= 12 OR           = 48

                    CP

            3•2•1• 2 •1= 12 OR

                       C  P

           

OR YOU CAN SOLVE IT THIS WAY:

CP= 1 entity = 2!       3 friends+CP= 4!

2!4!= 48

    c) If Pk and Christy cannot stand together, how many possibilities are there

          total arrangement- PK and Christy together= Christy and Pk not together

            120-48= 72

     d) What if JJ (J), Christy (C) and PK (P) must stand together, how many arrangements are possible

 JCP= 1 entity= 3!

2 friends= 2 entities

3!3!= 36

OR YOU CAN DO IT THIS WAY:

3•2•1•2•1=12 OR

JCP

2•3•2•1•1= 12 OR     =36

    JCP

2•1•3•2•1=12 

        JCP

Permutations with Case Restrictions:

1) How many different 4-digit numbers greater than 400 can you make using the digits 1, 2, 3, 4, 5 and 6. no digits are repeated

     1 •5 •4 •3= 60

     4

     1 •5 •4 •3= 60           =180

      5

    1 •5 •4 •3= 60

      6

OR YOU CAN DO IT THIS WAY:

3•5•4•3=180

4,5,6

I HOPE YOU ENJOYED MY BLOG POST. HAVE A GREAT DAY!

     

I CHOOSE!!!!!!!