##### Permutation and Combination

Concept | Formula | Explanation |
---|---|---|

Permutation (without repetition) | P(n, r) = n! / (n – r)! | Calculates the number of ways to arrange r items from a set of n items, without repetition. |

Permutation (with repetition) | P(n1, r1) * P(n2, r2) * … * P(nk, rk) | Calculates the number of ways to arrange items with repetitions, where n1, n2, …, nk are the number of distinct items and r1, r2, …, rk are the number of times each item can be repeated. |

Combination (without repetition) | C(n, r) = n! / (r! * (n – r)!) | Determines the number of ways to select r items from a set of n items, without considering the order. |

Combination (with repetition) | C(n + r – 1, r) = (n + r – 1)! / (r! * (n – 1)!) | Computes the number of ways to select r items from a set of n items, allowing repetitions. |

##### Factorial

###### Q1: Simplify 6!

Solution:

We can use the definition of factorial to simplify

6!:

6! = 6 x 5 x 4 x 3 x 2 x 1 = 720

###### b) Evaluate 9!/7!

We can simplify 9!/7! by dividing 9! by 7!

###### 9!/7! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 / (7 x 6 x 5 x 4 x 3 x 2 x 1) = 9 x 8 = 72

###### c) Evaluare 0!

By definition, 0! is equal to 1.

Therefore, 0! equals 1.

##### Permutation

###### Q1

a) How many different 4-digit numbers can be formed using the digits 1, 2, 3, and 4 without repeating any digit?

###### Solution: Since order matters in this case, we need to use the permutation formula:

###### P(4,4) = 4! = 24

###### Therefore, there are 24 different 4-digit numbers that can be formed using the digits 1, 2, 3, and 4 without repeating any digit.

###### b) A race has 5 runners. In how many ways can they finish first, second, and third?

###### Solution: Again, order matters in this case, so we can use the permutation formula:

###### P(5,3) = 5! / (5-3)! = 60

###### Therefore, there are 60 ways for the 5 runners to finish in first, second, and third place.

###### c) A company has 10 employees, including 2 managers, 3 supervisors, and 5 regular employees. In how many ways can a team of 4 employees be formed if it must include at least 1 supervisor?

###### Solution: We can use the permutation formula for this problem, but first we need to consider the possible combinations of employees:

###### 1 supervisor and 3 regular employees

2 supervisors and 2 regular employees

3 supervisors and 1 regular employee

###### For each combination, we can calculate the number of ways to choose the employees using the permutation formula:

###### 1 supervisor and 3 regular employees: P(3,3) x P(5,1) = 60

2 supervisors and 2 regular employees: P(3,2) x P(5,2) = 180

3 supervisors and 1 regular employee: P(3,1) x P(5,3) = 150

###### The total number of ways to form a team of 4 employees with at least 1 supervisor is the sum of these three values:

###### 60 + 180 + 150 = 390

###### Therefore, there are 390 ways to form a team of 4 employees

###### Q2: In a group of 8 people, how many ways are there to choose a president, vice president, and treasurer?

Solution: Using the formula for permutations, we have:

###### P(8,3) = 8! / (8-3)!

= 8! / 5!

= 8 x 7 x 6

= 336

###### However, we don’t care about the order in which the people are chosen, only who is assigned to each role. So we need to divide by the number of ways the 3 people can be rearranged (i.e. by 3!):

###### 336 / 3! = 336 / 6 = 56

###### Therefore, there are 56 ways to choose a president, vice president, and treasurer from a group of 8 people.

###### Q3: In how many ways can 5 books be arranged on a bookshelf?

Solution: Since the order matters, we can use the permutation formula. The number of ways to arrange 5 books on a bookshelf is 5P5 = 5! / (5-5)! = 120.

###### Q4: In how many ways can a committee of 4 be chosen from 10 people?

Solution: Again, since the order matters, we can use the permutation formula. The number of ways to choose a committee of 4 from 10 people is 10P4 = 10! / (10-4)! = 5040.

###### Q5: In how many ways can 3 different books be selected from a shelf of 7 books if the order in which they are selected is important?

Solution: We can use the permutation formula for this problem as well. The number of ways to select 3 different books from a shelf of 7 books is 7P3 = 7! / (7-3)! = 210.

###### Q6: In how many ways can a team of 5 people be chosen from a group of 10 people?

Solution: Since the order does not matter, we can use the combination formula. The number of ways to choose a team of 5 people from a group of 10 people is 10C5 = 10! / (5! * (10-5)!) = 252.

###### Q7: In how many ways can 5 letters be selected from the letters A, B, C, D, E, and F, if the order in which they are selected is important?

Solution: We can use the permutation formula for this problem. The number of ways to select 5 letters from 6 is 6P5 = 6! / (6-5)! = 720.

###### Q8: In how many ways can 3 letters be selected from the letters A, B, C, D, E, and F, if the order in which they are selected is important?

Solution: Again, we can use the permutation formula. The number of ways to select 3 letters from 6 is 6P3 = 6! / (6-3)! = 120.

###### Q9: In how many ways can 3 letters be selected from the letters A, B, C, D, E, and F, if the order in which they are selected is not important?

Solution: Since the order does not matter, we can use the combination formula. The number of ways to select 3 letters from 6 is 6C3 = 6! / (3! * (6-3)!) = 20.

###### Q10: In how many ways can 5 students be chosen from a class of 20 to form a committee, if the order in which they are chosen is important?

Solution: We can use the permutation formula for this problem. The number of ways to choose a committee of 5 students from a class of 20 is 20P5 = 20! / (20-5)! = 1,860,480.

###### Q11: In how many ways can a team of 3 be chosen from 7 girls and 5 boys?

Solution: Since the order does not matter, we can use the combination formula. The number of ways to choose a team of 3 from 7 girls and 5 boys is 12C3 = 12! / (3! * (12-3)!) = 220.

###### Q12: In how many ways can a committee of 4 officers (president, vice-president, secretary, treasurer) be chosen from a group of 10 people?

Solution:

Using the permutation formula, we have P(10,4) = 10! / (10-4)! = 10 x 9 x 8 x 7 = 5,040 ways.

###### Q13: In how many ways can 3 books be selected from a shelf containing 10 books if the order in which they are selected matters?

Solution:

Using the permutation formula, we have P(10,3) = 10! / (10-3)! = 10 x 9 x 8 = 720 ways.

###### Q14: A 6-digit number is formed using the digits 1, 2, 3, 4, 5, and 6 without repetition. What is the probability that the number is divisible by 3?

Solution:

There are 6! = 720 possible arrangements of the digits. To be divisible by 3, the sum of the digits must be divisible by 3. The sum of the digits 1 to 6 is 21, which is divisible by 3. There are two possible ways to arrange the digits to form a number divisible by 3: 123456 and 654321. So the probability is 2/720 = 1/360.

###### Q15: In how many ways can 5 people be arranged in a row for a photograph?

Solution:

Using the permutation formula, we have P(5,5) = 5! = 120 ways.

##### Combination

combination is a selection of items from a larger collection, where the order in which the items are chosen does not matter. The formula for calculating the number of combinations of r items chosen from a collection of n items is:

C(n, r) = n! / (r!(n-r)!)

where n! represents the factorial of n, or the product of all positive integers up to and including n.