Question

38. A State issues car license plates with three letters followed by three numbers (for example, FGH831 or BBB222). How many different license plates are possible? (Note that letters or numbers may repeat in the license plate. Hint: Use the Fundamental Counting Rule for this question.) There are five students who are interested in presenting their final project to the class, but there is only time for three presentations. The five students are Amy, Bob, Chun, Dan and Ed. 39. List all the possible combinations of three students the professor could pick. The order of the presentations does not matter. To save space, you can just use the first initial of each name if you like. 40. How many different combinations are there? 41. Now, use the Counting Rule for Combinations formula to calculate the number of combinations and compare your results. 42. Which approach was easier? 43. If the professor picks three students at random by drawing three names from a hat, what is the probability that the three students chosen are Amy, Chun and Dan? Now, assume that the first student chosen will present on Monday, the second will present on Wednesday and the third will present on Friday. In other words, now the order the students are chosen matters. Use this situation to answer the next three questions. 44. If the professor chooses Amy, Chun and Dan to present, list all the possible permutations of how their presentations could be distributed over the three days. 45. Use the Counting Rule for Permutations formula to calculate the total number of permutations when the professor is picking three students from the original group of five and the order matters. 46. If the professor picks three students at random from the original group of five, what is the probability that the results will be that Bob will present on Monday, Ed will present on Wednesday and Chun will present on Friday? Give your answer as a fraction.

Answer 1

I will be answering the first 4 questions (38,39,40,41) as there are more than 4 :

38.

no. of different possible plates = (no. of choice for 1st letter)*(no. of choice for 2nd letter)*(no. of choice for 3rd letter)*(no. of choice for 1st number)*(no. of choice for 2nd number)*(no. of choice for 3rd number)

no. of different possible plates = 26*26*26*10*10*10

no. of different possible plates = 17576000

39.

Amy=A, Bob=B, Chun=C, Dan=D and Ed=E

all the possible combinations :

ABC, ABD,ABE,ACD,ACE.ADE,BCD,BCE,BDE,CDE

40.

ABC, ABD,ABE,ACD,ACE.ADE,BCD,BCE,BDE,CDE

on counting them we can see :

there are 10 combinations

41.

using the Counting Rule for Combinations formula

no. of combinations = 5C3 = 5! / (3! * 2!) = 10

the results are same in both listing and counting rule of combinations method.

P.S. (please upvote if you find the answer satisfactory)