Permutations with case restriction

Last Friday, we continued discussing about permutations with restrictions to consider.

Remember that circular permutation DOES NOT HAVE first or last position; also, the number of permutation of n objects in a circle is (n-1)!

EXAMPLE 1.

a)      In how many 4  “words” are possible using the letters in MONDAY?

6 • 5 • 4 • 3 = 360

b) how many 4 letter "words" are possible using the letters in MONDAY if A is the third letter?

5 • 4 • 1 • 3 = 60

                                               A

c) how many 6 letter "words" are possible using the letters in GADGET?

6 • 5 • 4 • 3 • 2 • 1 = 720

720 ÷ 2 =  360

 d) how may 6 letter "words"  ending with G are possible using the letters in GADGET? 

5 • 4 • 3 • 2 • 1 • 2 = 240

                                                    G

240 ÷ 2 = 120

****Always remember divide by 2 the repeated items.

EXAMPLE 2. 

Three sets of book are being arranged on a shelf. the first set has 5 volumes, the second set has 3 volumes, and the third set has 2 volumes. In how many ways can the book be arranged if the volumes of each set are to be kept together?

 (5!)(3!)(2!)(3!) = 8640

5! - 1st set

3!- second set

2! - 3rd set

3! - 3 units

EXAMPLE 3. 

How many ways can 6 people be seated around a circular table if Georgia and Katie sit together but Eddie and Chris refuse to be seated together?

1. (n-1)!  >  (6-1)! = 5!  = 120      -----> K & G not together.  REMEMBER THAT G & K MUST BE TOGETHER.
2. (5-1)! = 4! > ( 4! ) ( 2! ) = 48 (total) ------> the 4 other people. K & G together.
3. (4-1)! = 3! > (3!) (2!) (2!) = 24 -----> E & C together. 
4. total - E & C together >  48 - 24 = 24