permutations and combinations review
name:
1. how many arrangements of the word curriculum can be made by using all of the letters each time?
2. how many arrangements can be made of word chemistry if the letters m, i, and s must appear together, in that order?
3. how many different license plates can be made using 1 letter followed by 5 digits if repetitions are allowed?
4. using the digits 0,7,6,9, determine how many 3 - digit odd numbers can be created if repetitions are allowed?
5. seven people get on a bus that has 3 empty seats. in how many ways can they seat themselves?
6. a football league consists of 7 teams. how many games are required if each team plays each other 5 times?
7. in how many ways can 3 different science books and 4 different mathematics books be arranged on a shelf so that the 3 science books will always be together?
8. how many ways can 5 different books be arranged on a shelf if two specified books must not be side by side?
9. determine the number of 4 - letter words which can be made from the letters of the word clover if the letter v must be included.
10. a 6 - player volleyball team stands in a straight line for a picture. if two particular players, joan and emily, must be together, then how many different arrangements can be made for the picture?
11. solve for n in the following equations algebraically
a) (_{n}p_{2}=6)
b) (_{n}p_{3}=n(n - 1))

Question

Accepted Answer

### 1. How many arrangements of the word CURRICULUM can be made by using all of the letters each time?
# Explanation:
## Step1: Count letter frequencies
The word CURRICULUM has 10 letters with 2 C's, 2 R's, 2 U's.
The formula for permutations of a word with repeated - letters is $\frac{n!}{n_1!n_2!\cdots n_k!}$, where $n$ is the total number of letters and $n_i$ are the number of times each repeated letter appears. Here $n = 10$, $n_1=n_2=n_3 = 2$.
## Step2: Calculate the number of arrangements
$\frac{10!}{2!2!2!}=\frac{10	imes9	imes8	imes7	imes6	imes5	imes4	imes3	imes2	imes1}{(2	imes1)(2	imes1)(2	imes1)}=\frac{3628800}{8}=453600$
# Answer:
453600

### 2. How many arrangements can be made of word CHEMISTRY if the letters M, I, and S must appear together, in that order?
# Explanation:
## Step1: Treat "MIS" as one unit
Treat "MIS" as a single entity. Then the number of entities to arrange is $9 - 3+1=7$ (since we combined 3 letters into 1).
The number of permutations of $n$ distinct objects is $n!$. Here $n = 7$.
## Step2: Calculate the number of arrangements
$7!=7	imes6	imes5	imes4	imes3	imes2	imes1 = 5040$
# Answer:
5040

### 3. How many different license - plates can be made using 1 letter followed by 5 digits if repetitions are allowed?
# Explanation:
## Step1: Determine the number of choices for the letter
There are 26 letters in the alphabet, so there are 26 choices for the letter.
## Step2: Determine the number of choices for the digits
Since repetitions are allowed and there are 10 digits (0 - 9), for each of the 5 digit - positions, there are 10 choices. By the multiplication principle, the number of ways to choose 5 digits is $10	imes10	imes10	imes10	imes10=10^5$.
## Step3: Calculate the total number of license - plates
By the multiplication principle, the total number of license - plates is $26	imes10^5 = 2600000$
# Answer:
2600000

### 4. Using the digits 0, 7, 6, 9, determine how many 3 - digit odd numbers can be created if repetitions are allowed?
# Explanation:
## Step1: Determine the units digit
For a number to be odd, the units digit can be 7 or 9. So there are 2 choices for the units digit.
## Step2: Determine the hundreds and tens digits
Since repetitions are allowed and the hundreds digit cannot be 0 (because then it would be a 2 - digit number), for the hundreds digit, there are 3 choices (7, 6, 9 if the units digit is 7 or 9; we exclude 0). For the tens digit, there are 4 choices (0, 7, 6, 9).
## Step3: Calculate the total number of 3 - digit odd numbers
By the multiplication principle, the total number of 3 - digit odd numbers is $3	imes4	imes2=24$
# Answer:
24

### 5. Seven people get on a bus that has 3 empty seats. In how many ways can they seat themselves?
# Explanation:
## Step1: Identify the permutation problem
This is a permutation problem of choosing 3 people out of 7 to sit in the 3 seats. The formula for permutations is $_{n}P_{r}=\frac{n!}{(n - r)!}$, where $n = 7$ and $r = 3$.
## Step2: Calculate the number of ways
$_{7}P_{3}=\frac{7!}{(7 - 3)!}=\frac{7!}{4!}=\frac{7	imes6	imes5	imes4!}{4!}=7	imes6	imes5 = 210$
# Answer:
210

### 6. A football league consists of 7 teams. How many games are required if each team plays each other 5 times?
# Explanation:
## Step1: Calculate the number of pairs of teams
The number of ways to choose 2 teams out of 7 for a game (using the combination formula $_{n}C_{r}=\frac{n!}{r!(n - r)!}$, where $n = 7$ and $r = 2$) is $_{7}C_{2}=\frac{7!}{2!(7 - 2)!}=\frac{7	imes6	imes5!}{2	imes1	imes5!}=21$.
## Step2: Calculate the total number of games
Since each pair of teams plays 5 times, the total number of games is $21	imes5 = 105$
# Answer:
105

### 7. In how many ways can 3 different science books and 4 different mathematics books be arranged on a shelf so that the 3 science books will always be together?
# Explanation:
## Step1: Treat the 3 science books as one unit
Treat the 3 science books as a single entity. Then the number of entities to arrange is $4 + 1=5$ (4 math books and 1 group of science books).
The number of permutations of these 5 entities is $5!$.
The number of permutations of the 3 science books within the group is $3!$.
## Step2: Calculate the total number of arrangements
By the multiplication principle, the total number of arrangements is $5!	imes3!=120	imes6 = 720$
# Answer:
720

### 8. How many ways can 5 different books be arranged on a shelf if two specified books must NOT be side - by - side?
# Explanation:
## Step1: Calculate the total number of arrangements of 5 books
The total number of arrangements of 5 different books is $5!=120$.
## Step2: Calculate the number of arrangements where the two specified books are together
Treat the two specified books as one unit. Then the number of entities to arrange is $4$ (the combined unit and the other 3 books). The number of permutations of these 4 entities is $4!$, and the number of permutations of the two specified books within the unit is $2!$. So the number of arrangements where the two specified books are together is $4!	imes2!=48$.
## Step3: Calculate the number of arrangements where the two specified books are not together
Subtract the number of arrangements where the two specified books are together from the total number of arrangements. So it is $5!-4!	imes2!=120 - 48=72$
# Answer:
72

### 9. Determine the number of 4 - letter words which can be made from the letters of the word CLOVER if the letter V must be included.
# Explanation:
## Step1: Calculate the number of ways to choose the remaining 3 letters
The word CLOVER has 6 letters. We must include V. So we need to choose 3 letters out of the remaining 5 letters. The number of ways to do this is $_{5}C_{3}=\frac{5!}{3!(5 - 3)!}=\frac{5	imes4	imes3!}{3!	imes2	imes1}=10$.
## Step2: Calculate the number of permutations of the 4 - letter combinations
For each combination of 4 letters (with V included), the number of permutations of 4 letters is $4!$.
## Step3: Calculate the total number of 4 - letter words
By the multiplication principle, the total number of 4 - letter words is $_{5}C_{3}	imes4!=10	imes24 = 240$
# Answer:
240

### 10. A 6 - player volleyball team stands in a straight line for a picture. If two particular players, Joan and Emily, must be together, then how many different arrangements can be made for the picture?
# Explanation:
## Step1: Treat Joan and Emily as one unit
Treat Joan and Emily as a single entity. Then the number of entities to arrange is $6 - 2 + 1=5$ (since we combined 2 players into 1).
The number of permutations of these 5 entities is $5!$.
The number of permutations of Joan and Emily within the unit is $2!$.
## Step2: Calculate the total number of arrangements
By the multiplication principle, the total number of arrangements is $5!	imes2!=120	imes2 = 240$
# Answer:
240

### 11. Solve for $n$ in the following equations algebraically:
#### a) $_{n}P_{2}=6$
# Explanation:
## Step1: Write the permutation formula
The formula for $_{n}P_{r}=\frac{n!}{(n - r)!}$. For $r = 2$, $_{n}P_{2}=\frac{n!}{(n - 2)!}=n(n - 1)$.
## Step2: Set up the equation and solve
We have the equation $n(n - 1)=6$, which expands to $n^{2}-n - 6=0$.
Factor the quadratic equation: $(n - 3)(n + 2)=0$.
So $n=3$ or $n=-2$. But since $n$ represents the number of objects and must be a non - negative integer, $n = 3$.
# Answer:
$n = 3$

#### b) $_{n}P_{3}=n(n - 1)$
# Explanation:
## Step1: Write the permutation formula
The formula for $_{n}P_{r}=\frac{n!}{(n - r)!}$. For $r = 3$, $_{n}P_{3}=\frac{n!}{(n - 3)!}=n(n - 1)(n - 2)$.
## Step2: Set up the equation and solve
We have the equation $n(n - 1)(n - 2)=n(n - 1)$.
First, assume $n(n - 1)
eq0$, then we can divide both sides by $n(n - 1)$ to get $n-2 = 1$, so $n=3$.
If $n(n - 1)=0$, then $n = 0$ or $n = 1$. But for the permutation $_{n}P_{3}$, $n\geq3$. So the solution is $n = 3$.
# Answer:
$n = 3$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

1. How many arrangements of the word CURRICULUM can be made by using all of the letters each time?

Step1: Count letter frequencies

Step2: Calculate the number of arrangements

Step1: Treat "MIS" as one unit

Step2: Calculate the number of arrangements

Step1: Determine the number of choices for the letter

Step2: Determine the number of choices for the digits

Step3: Calculate the total number of license - plates

Answer:

2. How many arrangements can be made of word CHEMISTRY if the letters M, I, and S must appear together, in that order?