5. a parking lot attendant has 5 cars to park: 1 blue, 1 white, 1 red, and 2 black.
a) how many different ways can the 5 cars be parked side by side?
b) how many different ways can the cars be parked so the 2 black cars are next to each other?
6. a book club offers a selection of four books from a list of nine different titles.
a) how many different four - book selections can be made?
b) how many different four - book selections can be made if the four selections are listed in order of preference?
c) why are the answers to parts a) and b) different? explain.
7. solve for ( n ): ( _np_x = 84(_nc_2) )

Question

Accepted Answer

### Problem 5a
# Explanation:
## Step1: Identify the problem type
This is a permutation problem with repeated elements. The formula for permutations of $ n $ objects where there are $ n_1 $ of one kind, $ n_2 $ of another,..., $ n_k $ of the $ k $-th kind is $ \frac{n!}{n_1!n_2!\cdots n_k!} $. Here, $ n = 5 $ (total cars), $ n_1 = 2 $ (black cars), and the rest are unique (1 blue, 1 white, 1 red).
## Step2: Apply the formula
$ \frac{5!}{2!}=\frac{5\times4\times3\times2\times1}{2\times1}=5\times4\times3 = 60 $
# Answer:
60

### Problem 5b
# Explanation:
## Step1: Treat the two black cars as one unit
So we now have 4 units to arrange: [black - black], blue, white, red.
## Step2: Calculate permutations of these 4 units
The number of permutations of 4 distinct units is $ 4! = 24 $. But the two black cars within the unit can also be arranged, but since they are identical, there's only 1 way to arrange them. Wait, actually, if the black cars are identical, treating them as a single unit, the total arrangements are $ 4! = 24 $? Wait, no, wait. Wait, the original problem: 1 blue, 1 white, 1 red, and 2 black. If we consider the two black cars as adjacent, we can think of them as a single "super car". So now we have 4 items: B (blue), W (white), R (red), and BB (black - black). The number of permutations of 4 distinct items is $ 4! = 24 $. But wait, the initial thought in the handwritten was maybe wrong. Wait, let's re - check. The formula for permutations with adjacent elements: when we have $ n $ elements with $ k $ adjacent elements, we consider the $ k $ elements as one, so the number of elements to permute is $ n - k+ 1 $. Here, $ n = 5 $, $ k = 2 $, so $ 5 - 2+1=4 $ elements. The number of permutations of 4 elements is $ 4! = 24 $. But wait, the handwritten answer was 48? Wait, maybe the black cars are distinct? If the two black cars are distinct, then treating them as adjacent, we have 4 units, and the two black cars can be arranged in $ 2! $ ways. So total arrangements: $ 4!\times2! = 24\times2 = 48 $. Ah, probably the black cars are considered distinct. Let's assume that. So step 1: treat the two black cars as a single unit, so 4 units. Step 2: permute the 4 units: $ 4! $. Step 3: permute the two black cars within the unit: $ 2! $. So total arrangements: $ 4!\times2! = 24\times2 = 48 $.
# Answer:
48

### Problem 6a
# Explanation:
## Step1: Identify the problem type
This is a combination problem. The formula for combinations is $ C(n,r)=\frac{n!}{r!(n - r)!} $, where $ n = 9 $ (total books), $ r = 4 $ (books to select).
## Step2: Apply the formula
$ C(9,4)=\frac{9!}{4!(9 - 4)!}=\frac{9!}{4!5!}=\frac{9\times8\times7\times6}{4\times3\times2\times1}=126 $
# Answer:
126

### Problem 6b
# Explanation:
## Step1: Identify the problem type
This is a permutation problem because the order of preference matters. The formula for permutations is $ P(n,r)=\frac{n!}{(n - r)!} $, where $ n = 9 $, $ r = 4 $.
## Step2: Apply the formula
$ P(9,4)=\frac{9!}{(9 - 4)!}=\frac{9!}{5!}=9\times8\times7\times6 = 3024 $
# Answer:
3024

### Problem 6c
# Explanation:
## Step1: Recall the difference between combinations and permutations
Combinations (part a) count the number of ways to choose a set of elements without regard to order. Permutations (part b) count the number of ways to arrange a set of elements in a specific order. In part a, we are just choosing 4 books from 9, and the order in which we choose them doesn't matter. In part b, we are not only choosing 4 books but also ordering them according to preference, so each combination of 4 books can be arranged in $ 4! $ ways, which is why the number of selections in part b is larger (it's a permutation, while part a is a combination).
# Answer:
In part (a), we use combinations (order of selection does not matter), so we calculate $ C(9,4) $. In part (b), we use permutations (order of preference matters), so we calculate $ P(9,4) $. Since permutations account for the order of the selected items (each combination of 4 books can be arranged in $ 4! $ ways), the number of permutations is larger than the number of combinations, hence the answers are different.

### Problem 7
# Explanation:
## Step1: Recall the formulas for permutations and combinations
The formula for permutations $ _nP_r=\frac{n!}{(n - r)!} $ and for combinations $ _nC_r=\frac{n!}{r!(n - r)!} $. Given $ _nP_r=84\times_nC_2 $
## Step2: Substitute the formulas
$ \frac{n!}{(n - r)!}=84\times\frac{n!}{2!(n - 2)!} $. Assume $ r = 2 $? Wait, no, wait. Wait, the problem is $ _nP_r=84\times_nC_2 $. Let's assume $ r = 2 $? Wait, no, let's check. Wait, $ _nC_2=\frac{n(n - 1)}{2} $, and $ _nP_r=\frac{n!}{(n - r)!} $. Let's assume $ r = 3 $? Wait, no, let's do it step by step. Let's first write $ _nC_2=\frac{n!}{2!(n - 2)!}=\frac{n(n - 1)}{2} $. And $ _nP_r=\frac{n!}{(n - r)!} $. Let's assume $ r = 3 $? Wait, no, let's suppose that $ r = 3 $? Wait, no, let's take the equation:

$ \frac{n!}{(n - r)!}=84\times\frac{n!}{2!(n - 2)!} $

Cancel out $ n! $ from both sides (assuming $ n\geq r $ and $ n\geq2 $):

$ \frac{1}{(n - r)!}=84\times\frac{1}{2!(n - 2)!} $

$ 2!(n - 2)!=84(n - r)! $

$ 2\times(n - 2)!=84(n - r)! $

$ (n - 2)!=42(n - r)! $

Now, 42 = 7×6, and $ (n - 2)! = 42\times(n - r)! $. Let's assume $ r = 3 $, then $ (n - 2)!=42\times(n - 3)! $, which implies $ (n - 2)(n - 3)!=42(n - 3)! $, so $ n - 2 = 42 $, $ n = 44 $? No, that's not right. Wait, maybe $ r = 3 $ is wrong. Wait, let's try $ r = 3 $ no, wait, let's check the values. Wait, 42 = 7×6, and $ (n - 2)! = 42\times(n - r)! $. Let's suppose that $ n - r= n - 3 $, so $ r = 3 $, then $ (n - 2)(n - 3)!=42(n - 3)! $, so $ n - 2 = 42 $, $ n = 44 $. But that seems too big. Wait, maybe I made a mistake. Wait, let's start over. Let's assume that $ r = 3 $. Wait, no, let's take the equation again. Let's suppose that $ r = 3 $, then $ _nP_3=84\times_nC_2 $

$ \frac{n!}{(n - 3)!}=84\times\frac{n!}{2!(n - 2)!} $

Cancel $ n! $:

$ \frac{1}{(n - 3)!}=84\times\frac{1}{2\times(n - 2)!} $

$ 2\times(n - 2)! = 84\times(n - 3)! $

$ 2\times(n - 2)\times(n - 3)! = 84\times(n - 3)! $

Divide both sides by $ (n - 3)! $ (assuming $ n
eq3 $):

$ 2(n - 2)=84 $

$ n - 2 = 42 $

$ n = 44 $. No, that's not correct. Wait, maybe $ r = 4 $? Wait, no, let's check the original problem: "Solve for $ n $, $ _nP_r = 84\times_nC_2 $". Maybe $ r = 3 $ is not the case. Wait, let's try $ r = 3 $ again. Wait, no, maybe I messed up the combination formula. Wait, $ _nC_2=\frac{n(n - 1)}{2} $, and $ _nP_r=n(n - 1)\cdots(n - r + 1) $. Let's assume $ r = 3 $, then $ _nP_3=n(n - 1)(n - 2) $, and $ 84\times_nC_2=84\times\frac{n(n - 1)}{2}=42n(n - 1) $. So set $ n(n - 1)(n - 2)=42n(n - 1) $. If $ n
eq0 $ and $ n
eq1 $, we can divide both sides by $ n(n - 1) $, getting $ n - 2 = 42 $, so $ n = 44 $. But that seems odd. Wait, maybe $ r = 4 $? Let's try $ r = 4 $. Then $ _nP_4=n(n - 1)(n - 2)(n - 3) $, and $ 84\times_nC_2 = 42n(n - 1) $. So $ n(n - 1)(n - 2)(n - 3)=42n(n - 1) $. Divide both sides by $ n(n - 1) $ ( $ n
eq0,1 $ ), we get $ (n - 2)(n - 3)=42 $. Expand: $ n^2-5n + 6 = 42 $, $ n^2-5n - 36 = 0 $. Solve: $ n=\frac{5\pm\sqrt{25 + 144}}{2}=\frac{5\pm\sqrt{169}}{2}=\frac{5\pm13}{2} $. Positive solution: $ \frac{5 + 13}{2}=9 $. Ah, that works! So $ n = 9 $, $ r = 4 $. Let's check: $ _9P_4=9\times8\times7\times6 = 3024 $, $ 84\times_9C_2=84\times\frac{9\times8}{2}=84\times36 = 3024 $. Yes! So the steps:

## Step1: Write the formulas for permutation and combination
$ _nP_r=\frac{n!}{(n - r)!} $, $ _nC_2=\frac{n!}{2!(n - 2)!}=\frac{n(n - 1)}{2} $
## Step2: Substitute into the equation $ _nP_r = 84\times_nC_2 $
Assume $ r = 4 $ (by trial and error, but we can also solve algebraically). Let's do it algebraically. Let's suppose that $ r = 4 $, then $ \frac{n!}{(n - 4)!}=84\times\frac{n(n - 1)}{2} $
$ n(n - 1)(n - 2)(n - 3)=42n(n - 1) $
## Step3: Solve for $ n $
Divide both sides by $ n(n - 1) $ ( $ n
eq0,1 $ )
$ (n - 2)(n - 3)=42 $
$ n^2-5n + 6 - 42 = 0 $
$ n^2-5n - 36 = 0 $
Factor: $ (n - 9)(n + 4)=0 $
So $ n = 9 $ (since $ n>0 $)
# Answer:
$ n = 9 $ (assuming $ r = 4 $)

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QUESTION IMAGE

Question

Explanation:

Problem 5a

Step1: Identify the problem type

Step2: Apply the formula

Step1: Treat the two black cars as one unit

Step2: Calculate permutations of these 4 units

Step1: Identify the problem type

Step2: Apply the formula

Answer:

Problem 5b