5 Use the word ADDRESSES for the following a How many total

5. Use the word ADDRESSES for the following:

a. How many total permutations are there of the letters of the word ADDRESSES?

b. How many 8permutations are there of these nine letters?

c. How many 8permutations are there of these nine letters in a circular fashion?

6. Determine the number of 11permutations of the multiset S = {3 a, 3 b, 3 c, 3 d}.   

7. A bagel store sells six different kinds of bagels. Suppose you choose 15 bagels at random. What is the probability that your choice contains at least one bagel of each kind?

8. A bag contains 100 apples, 100 bananas, 100 oranges, and 100 pears. If I pick one piece of fruit out of the bag every minute, how long will it be before I am assured of having picked at least a dozen pieces of fruit of the same kind?

9. There are 250 students living a dorm. Can we be assured/guaranteed that there are 5 students who share a birthday in the same month? How do you know?

Solution

Permutations of 9 letters P(9, 9)

Permutations of the two ‘D’ letters P(2, 2)

Permutations of the two ‘E’ letters P(2, 2)

Permutations of the three ‘S’ letters P(3, 3)

answer = 9! / (2!*2!*3!)=15120

b)

8-permutations without a letter ‘A’ 8! / (2!*2!*3!)

8-permutations without a letter ‘D’ 8!/ (2!*3!)

8-permutations without a letter ‘R   8! / (2!*2!*3!)

8-permutations without a letter ‘E’     8!/ ( 2!*3!)

8-permutations without a letter ‘S’ 8! / (2!*2!*2!)

Therefore the answer is 2 * 8! / ( 2!*2!*3!) + 2* 8!/(2!*3!) + 8! / (2!*2!*2!)

15120 .

6.

There are three types of objects in the multiset S which has size 12. To get 11- permutation, one of the objects has be left behind. There are 4 different ways to leave one object behind, that is, either an a, or a b, or a c or a d

The number of 11-permutations with 2 a’s is 11! / ( 2! · 3! · 3! 3! )

The number of 11-permutations with 2 b’s is 11! / ( 2! · 3! · 3! 3! )

The number of 11-permutations with 2 c’s is 11! / ; ( 2! · 3! · 3! 3! )

The number of 11-permutations with 2 d’s is 11! / ( 2! · 3! · 3! 3! )

Thus the total number of 11-permutations is 4* [11! / ( 2! · 3! · 3! 3! )]

FOR THE OTHERS QUESTION

I CAN GLADLY HELP YOU BUT YOU SHOULD POST IT IN A NEW QUESTION.