(3 points) In how many ways can 3 diflerent novels, 4 different mathematics book
ID: 2946509 • Letter: #
Question
(3 points) In how many ways can 3 diflerent novels, 4 different mathematics books, and 1 biology book be arranged on a bookshelf i (a) the books can be arranged in any order? Answer (b) the mathematics books must be together and the novels must be together? Answer (c) the mathematics books must be together but the other books can be arranged in any order? Answer (2 points) How many different 10-letter permutations can be formed from 8 identical H's and two identical T's? Answer: (2 points) How many three-letter "words" can be made from 10 letters FGHIJKLMNO" if repetition of letters (a) is allowed? Your answer is: (b) is not allowed? Your answer is: (3 points) Find the value of each of the following quantitios C(11,4 C(6,6) C(8,8) (2 points) How many five card hands consisting of 2 kings and 3 aces can be dealt from a deck of 52 playing cards? Answer: 1 point) How many ways are there to select 12 countries in the United Nations to serve on a counoil i 3 is selected from a block of 52 3 are selected from a block of 67 and 6 are selected from the remaining 70 countries? (2 points) A school dance committee is to consist of 2 freshmen, 3 sophomores, 4 juniors, and 5 seniors. If 7 freshmen, 7 sophomores 7 juniors, and 9 seniors are eligible to be on the committee, in how many ways can the comminee be chosen? Your answer is:Explanation / Answer
(1)
Total books = 3+4+1 = 8
(a)
Total possible arrangements = 8! = 40320
(b)
Considering the maths books as a single group and novels as a single group, we have 3 groups now.
Total combinations = 3!
Accounting for the possible arrangements within these groups, we have:
Total possible combinations = 3!*3!*4! = 864
(c)
Considering the math books as a single group, we have 5 different entities.
Accounting for the possible arrangements within the maths group, we have:
Total possible combinations = 5!*4! = 2880
(2)
Total possible permutations are given by: 10!/(8!*2!) = 45
(3)
(a)
If repetition is allowed, then we have 10 choices for each of the three positions, so the total possible words = 10^3 = 1000
(b)
If repetition is not allowed, then the possible words are: 10P3 = 10!/((10-3)!) = 720
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