. How many ways could people rank the top three cows from a herd of ten. A) 210
ID: 2493366 • Letter: #
Question
. How many ways could people rank the top three cows from a herd of ten. A) 210 B) 720 C) 5,040 D) 120 33.
A business has seven locations to choose from and wishes to rank only the top three locations. How many different ways can this be done? A) 5,040 B) 210 C) 840 D) 420
34. How many different ways can four people: Andy, Betty, Cindy, and Doug, sit in a row at the opera if Andy and Betty must sit together? A) 24 B) 12 C) 6 D) 18
35. Evaluate the following: 7C3. A) 210 B) 5,040 C) 35 D) 6
36. How many ways can a student select five questions from an exam containing 12 questions, if one of the five must be the last question? A) 330 B) 7920 C) 40,320 D) 95,040
37. A certain system has two components. There are 8 different models of the first component and 9 different models of the second. Any first component can be paired with any second component. A salesman must select 2 of the first component and 3 of the second to take on a sales call. How many different sets of components can the salesman take?
38. 4 statistics professors and 6 chemistry professors are available to be advisors to a student organization. The student organization needs two of the professors to be advisors. If each professor has an equal chance of being selected, what is the probability that both professors are chemistry professors? A) 0.333 B) 0.267 C) 0.111 D) 0.100
Explanation / Answer
People rank the top three cows from a herd of ten = 10P3 = 10!/(10-3)! = 10*9*8=720
Answer : B) 720
A business has seven locations to choose from and wishes to rank only the top three locations. How many different ways can this be done?
Sollution: 7P3 =7!/(7-3)! = 7*6*5 = 210
Answer: B) 210
34.
Sitting arrangement of Andy, Betty, Cindy, and Doug, sit in a row at the opera if Andy and Betty must sit together =
Andy and Betty will be treated as 1 person.
So 3 people can sit 3! ways = 3*2*1 = 6
If Andy and Betty must sit together, they can be arranged = 2! = 2*1 = 2
The require number of sitting arrangement = 6*2 = 12
Answer:B) 12
35.
7C3 = 7!/(7-3)!*3! = (7*6*5)/(3*2*1) = 35
Answer: C) 35
36.
Student can choice 4 question out of 11 question. So ways are 11C4= 11!/(11-4)!*4!=(11*10*9*8)/(4*3*2*1)
=330
Answer: A) 330
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