My revised solution:b. 1c. 10d. 5e. 15

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This discussion focuses on combinatorial problems involving permutations and combinations using specific examples. The participant calculates the number of repeating and non-repeating three-digit numbers from the set {1, 2, 3}, yielding 27 and 6 respectively. Additionally, they analyze outcomes from tossing a fair coin four times, determining the number of ways to achieve various tail counts. The discussion also covers permutations in the context of a race with 12 runners and the calculation of unique four-digit ATM PINs without repeated digits, resulting in 5040 possible combinations.

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1. (a) How many repeating three-digit numbers can be formed using the numbers {1, 2, 3}?
(b) How many non-repeating three-digit numbers can be formed using the numbers {1, 2, 3}?

My solutions:
a. 3 x 3 x 3=27
b. 3 x 2 x 1 =6

2. (a) Construct a tree diagram for 4 tosses of a fair coin.
(b) How many ways can you get more than three tails?
(c) How many ways can you get fewer than three tails?
(d) How many ways can you get at least three tails?
(e) How many ways can you get no more than three tails?

My solutions:
b. 1
c. 12
d. 5
e. 15

3. Determine the number of possible ways to mark your answer sheet for each test:
(a) a six question true-or-false test.

My solutions:

a. 6 questions and two choices so 2 x 2 x 2 x 2 x 2 x 2= 64

6. Decide whether a permuation or a combination is needed:
(a) a telephone number
(b) a Social Security number
(c) a hand of cards in poker
(d) the combination on a student gym locker

My solutions:
a. permutation
b. permutation
c. combination
d. permutation

7. How many different ways could first, second, and third place finishers occur in a race with
12 runners competing?

My solution: 1st 2nd 3rd, 12 runners
(12
3 ) = 220 ways
n!/k! (n-k)!

8. An ATM requires a four-digit PIN number using the digits 0-9. How many such PINs
have no repeated digits?

My solution:
nPk= n!/n-k
10 numbers, 4 digits
(10
4)= 5040
 
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Hm, seems you might need to check 2c: there are sixteen possible outcomes; four different ways to have three tails, and one possible way to have four tails. Right?

I mean, clearly 2c + 2d = 16, so one of those answers must be wrong.
 

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