Mastering Factorials: Fun Quiz & Solutions!

A. Solve the following. Remember that the factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n. 0! is defined as 1.

 

Question 1: What is 6!?

 

Answer: 720

 

Solution:6! = 6 x 5 × 4 × 3 × 2 × 1 = 720

 

Question 2: Calculate 9!.

 

Answer: 362,880

 

Solution: 9! = 9 x 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362,880

 

Question 3: Simplify 10!/7!.

 

Answer: 90

 

Solution: 10!/7! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1/(7 × 6 × 5 × 4 × 3 × 2 × 1) = 10 × 9 x 8 = 720

 

Question 4: Evaluate 7!/(3!4!).

 

Answer: 35

 

Solution: 7!/(3!4!) = 7 × 6 × 5 × 4 × 3 × 2 × 1/{(3 × 2 × 1)(4 × 3 × 2 × 1)} = (7 × 6 × 5)/(3 × 2 × 1) = 7 × 5 = 35

 

Question 5: Solve for n: n! = 720

 

Answer: n = 6

 

Solution: We can test values: 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720. Therefore, n = 6.

 

Question 6: What is the value of 6!/(2!4!)?

 

Answer: 15

 

Solution: 6!/(2!4!) = 6 × 5 × 4 × 3 × 2 × 1/(2 × 1)(4 × 3 × 2 × 1)} = 6 × 5/2 = 15

 

Question 7: Simplify (n+2)!/n!.

 

Answer: (n+2)(n+1)

 

Solution: (n+2)!/n! = (n+2)(n+1)n!/n! = (n+2)(n+1)

 

Question 8: Find the value of 12!/(10!2!).

 

Answer: 66

 

Solution: 12!/(10!2!) = 12 × 11 × 10!/(10! × 2 × 1) = 12 × 11/2 = 6 × 11 = 66

 

Question 9: If n!/(n-2)! = 20, find n.

 

Answer: n = 5

 

Solution: n!/(n-2)! = n(n-1)(n-2)!/(n-2)! = n(n-1) = 20. Solving the quadratic equation n^2 - n - 20 = 0, we get (n-5)(n+4) = 0. Since n must be positive, n = 5.

 

Question 10: What is the value of 0!?

 

Answer: 1

 

Solution: By definition, 0! = 1. This is a convention established to maintain consistency in mathematical formulas and theorems involving factorials.

B. Solve the following Factorial word problems:

 

Question 1: A teacher wants to arrange 5 students in a row for a photograph. How many different arrangements are possible?

 

Answer: 120

 

Solution: This is a permutation problem. The number of arrangements of 5 students is 5! = 5 × 4 × 3 × 2 × 1 = 120.

 

Question 2: How many ways can you arrange the letters in the word "MATH"?

 

Answer: 24

 

Solution: There are 4 letters, so there are 4! = 4 × 3 × 2 × 1 = 24 ways to arrange them.

 

Question 3: A restaurant offers 7 different appetizers. If a customer orders 3 appetizers, how many different combinations are possible, assuming the order doesn't matter? (Note: This is a combination problem, not a direct factorial application, but it uses factorials in its solution)

 

Answer: 35

 

Solution: This uses combinations. The number of combinations is given by 7!/(3!4!) = 7 × 6 × 5/(3 × 2 × 1) = 35

 

Question 4: A bookshelf has space for 6 books. If you have 8 books, how many ways can you arrange 6 of them on the shelf?

 

Answer: 20160

 

Solution: This is a permutation. The number of ways is P(8,6) = 8!/(8-6)! = 8 × 7 × 6 × 5 × 4 × 3 = 20160.

 

Question 5: How many ways are there to arrange the letters of the word "APPLE"? (Consider that the letter P is repeated)

 

Answer: 60

 

Solution: There are 5 letters, but two are P's. The number of arrangements is 5!/2! = 120/2 = 60.

 

Question 6: A pizza shop offers 5 different toppings. How many different pizzas can be made with 3 toppings, if the order of toppings doesn't matter? (Combination)

 

Answer: 10

 

Solution: This is a combination. The number of combinations is 5!/(3!2!) = 5 × 4/2 = 10.

 

Question 7: Ten people are running a race. How many different ways can the first three places be awarded (gold, silver, bronze)?

 

Answer: 720

 

Solution: This is a permutation. The number of ways is P(10,3) = 10 × 9 × 8 = 720.

 

Question 8: A password must be 4 characters long, using only the digits 0-9. How many possible passwords are there if repetition is allowed?

 

Answer: 10000

 

Solution: There are 10 choices for each of the 4 positions, so there are 10⁴ = 10,000 possible passwords. (Not a direct factorial, but related to permutations with repetition).

 

Question 9: A club has 12 members. How many ways can they elect a president, vice-president, and treasurer?

 

Answer: 1320

 

Solution: This is a permutation. The number of ways is P(12,3) = 12 × 11 × 10 = 1320.

 

Question 10: How many different ways can 7 distinct books be arranged on a shelf?

 

Answer: 5040

 

Solution: This is a permutation of 7 items. The number of ways is 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040.

 

Remember that when order matters, it's a permutation; when order doesn't matter, it's a combination. Factorials are fundamental to calculating both.