Question 859654: Please help me solve this question. You can program some CD players to choose randomly in which order to play the tracks on a record.
a. How many different choices are there for a CD with 7 tracks?
b. Suppose you program the CD player to play 5 of the tracks, chosen randomly. In how many ways can the choice be made?
for a, I think the answer is 5040 but i am not sure. For b, I think the answer is 21 but I feel like the gap between the two numbers is too big.
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! You said the magic word. Please!
Q1 you have the correct answer - 5040. Here's how you get it.
You have 7 tracks, so how many can be your first choice? Seven is correct. For each of the seven, how many tracks can you choose for the secod track? Six is right! If you stopped now you would have 7*6 or 42 possible orders. Agree? But we want to continue. There are 5 choices for the third track, 4 for the fourth, 3 for the fifth, 2 for the sixth and 1 for the seventh. So the answer is simply
(1) Number of distinct choices of all 7 = 7*6*5*4*3*2*1 = 7! = 5040. The symbol ! is called the factorial, it's on your calculator. Enter 7, then hit the ! key and read 5040. If you don't have a factorial key on your calculator or not allowed to use a calculator, just multiply 7*6*5... , it's not hard or long.
Q2 is the same as Q1 except you stop at fifth track in the above sequence and get
(2) Number of distinct choices of 5 of the 7 = 7*6*5*4*3 = 7!/2! = 5040/2 = 2520. You're right to sense an error to 21.
Note that answer for Q2 has 5 numbers in the product. That's the easy way to get the answer. For example in the above where I got 7*6 after selecting 2 tracks is the product of two numbers. If you want to use 4 tracks the answer is 7*6*5*4.
There is a fancy name given to the above solutions. It's called the number of Permutations, given by the uppercase P. The P is written with a subscripted number in front and behind the P. I can't type the subs but it looks like this
(3) Number of distinct choices of all 7 = 7P7 = 7*6*5*4*3*2*1 = 7! = 5040.
and the
(4) Number of distinct choices of 5 of the 7 = 7P5 = 7*6*5*4*3 = 7!/2! = 5040/2 = 2520
If you have n items to choose from and you want to choose r of them the number of r groupings is
(5) nPr = n!/(n-r)!
Note; Permutations gives us distint groupings, wherein we say that "order matters".
Use (5) to solve your Q2 and get
(6) nPr = 7P5 = 7!/(7-5)! = 7!/2! = (7*6*5*4*3*2*1)/(2*1) = 7*6*5*4*3 = 5040
Note how the factorial in the denominator cancels the "tail end" of the numerator.
Let's do "How many ways can we select 3 students in a class of 20 to be the President, veep and secretary?"
Answer: 20P3 = 20*19*18 = 6840
Easy? Thank you for reading.
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