# Counting, Permutations, and Combinations - Grade 9 Worksheet
Section A: Multiple Choice Questions
Q1: A restaurant offers 4 appetizers, 6 main courses, and 3 desserts. How many different three-course meals can be ordered? (a) 13 (b) 24 (c) 72 (d) 108
Solution:
Ans: (c) Explanation: Using the Fundamental Counting Principle, multiply the number of choices for each course: \(4 \times 6 \times 3 = 72\). Option (a) incorrectly adds the choices. Option (b) represents only two categories multiplied. Option (d) uses an incorrect multiplication.
Q2: How many different ways can the letters in the word "MATH" be arranged? (a) 4 (b) 12 (c) 24 (d) 16
Solution:
Ans: (c) Explanation: The number of permutations of 4 distinct letters is \(4! = 4 \times 3 \times 2 \times 1 = 24\). Option (a) represents only the number of letters. Option (b) is \(4 \times 3\), missing the remaining factors. Option (d) represents \(4^2\), which is not the correct formula.
Q3: In how many ways can 3 students be selected from a group of 8 students to form a committee? (a) 24 (b) 56 (c) 336 (d) 512
Solution:
Ans: (b) Explanation: This is a combination problem since order doesn't matter. Use \(C(8,3) = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\). Option (a) represents \(P(4,3)\). Option (c) represents \(P(8,3)\), which accounts for order. Option (d) represents \(8^3\), which is incorrect.
Q4: A lock has a 4-digit code where each digit can be any number from 0 to 9. How many different codes are possible if digits can be repeated? (a) 40 (b) 5,040 (c) 10,000 (d) 1,000
Solution:
Ans: (c) Explanation: Using the Fundamental Counting Principle with repetition allowed: \(10 \times 10 \times 10 \times 10 = 10^4 = 10,000\). Option (a) incorrectly adds 4 and 10. Option (b) represents \(10 \times 9 \times 8 \times 7\) without repetition. Option (d) represents \(10^3\), using only 3 digits.
Q5: How many permutations are there of the letters in the word "BOOK"? (a) 24 (b) 12 (c) 6 (d) 4
Solution:
Ans: (b) Explanation: The word "BOOK" has 4 letters with O repeated twice. Use the formula for permutations with repetition: \(\frac{4!}{2!} = \frac{24}{2} = 12\). Option (a) represents \(4!\) without accounting for repetition. Option (c) represents \(3!\). Option (d) is just the number of letters.
Q6: In how many ways can 5 people stand in a line? (a) 5 (b) 25 (c) 60 (d) 120
Solution:
Ans: (d) Explanation: The number of permutations of 5 people is \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Option (a) is just the count of people. Option (b) represents \(5^2\). Option (c) represents \(5 \times 4 \times 3\), which is incomplete.
Q7: How many ways can you choose 2 books from a shelf of 7 books? (a) 14 (b) 21 (c) 42 (d) 49
Solution:
Ans: (b) Explanation: This is a combination since order doesn't matter: \(C(7,2) = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21\). Option (a) represents \(7 \times 2\). Option (c) represents \(P(7,2) = 7 \times 6 = 42\), which counts order. Option (d) represents \(7^2\).
Q8: A student must answer 4 questions out of 6 questions on a test. In how many ways can the student select the questions? (a) 10 (b) 15 (c) 20 (d) 24
Solution:
Ans: (b) Explanation: This is a combination problem: \(C(6,4) = \frac{6!}{4!(6-4)!} = \frac{6 \times 5}{2 \times 1} = 15\). Note that \(C(6,4) = C(6,2)\). Option (a) represents \(C(5,2)\). Option (c) represents an incorrect calculation. Option (d) represents \(4!\).
Section B: Fill in the Blanks
Q9: The Fundamental Counting Principle states that if one event can occur in \(m\) ways and another event can occur in \(n\) ways, then both events can occur in __________ ways.
Solution:
Ans: \(m \times n\) Explanation: The Fundamental Counting Principle uses multiplication to find the total number of combined outcomes.
Q10: A permutation is an arrangement of objects where __________ matters.
Solution:
Ans: order Explanation: In a permutation, the sequence or arrangement of objects is important, unlike in combinations.
Q11: The notation \(n!\) is called __________ and means the product of all positive integers from 1 to \(n\).
Solution:
Ans: factorial Explanation: The factorial symbol \(n!\) represents \(n \times (n-1) \times (n-2) \times ... \times 2 \times 1\).
Q12: The number of permutations of \(n\) objects taken \(r\) at a time is denoted by __________ or \(_nP_r\).
Solution:
Ans: \(P(n,r)\) Explanation: The permutation notation \(P(n,r)\) or \(_nP_r\) represents arrangements of \(r\) objects chosen from \(n\) objects.
Q13: The formula for combinations is \(C(n,r) = \frac{n!}{r! \times __________}\).
Solution:
Ans: \((n-r)!\) Explanation: The combination formula is \(C(n,r) = \frac{n!}{r!(n-r)!}\), which calculates selections where order doesn't matter.
Q14: By definition, \(0!\) equals __________.
Solution:
Ans: 1 Explanation: By mathematical convention, \(0! = 1\). This definition is essential for formulas involving permutations and combinations.
Section C: Word Problems
Q15: A pizza shop offers 3 types of crust, 5 types of sauce, and 8 types of toppings. If a customer must choose one of each, how many different pizza combinations are possible?
Solution:
Ans: Using the Fundamental Counting Principle: \(3 \times 5 \times 8 = 120\) Final Answer: 120 different pizza combinations
Q16: A committee of 4 people needs to be formed from a group of 10 people. How many different committees can be formed?
Solution:
Ans: This is a combination problem since order doesn't matter: \(C(10,4) = \frac{10!}{4!(10-4)!} = \frac{10!}{4! \times 6!}\) \(= \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}\) \(= \frac{5040}{24} = 210\) Final Answer: 210 different committees
Q17: How many different 5-letter arrangements can be made using the letters in the word "LEVEL"?
Solution:
Ans: The word "LEVEL" has 5 letters: L appears 2 times, E appears 2 times, V appears 1 time. Using the formula for permutations with repetition: \(\frac{5!}{2! \times 2! \times 1!} = \frac{120}{2 \times 2 \times 1} = \frac{120}{4} = 30\) Final Answer: 30 different arrangements
Q18: A teacher needs to arrange 6 students in a row for a photograph. In how many different ways can this be done?
Solution:
Ans: This is a permutation of 6 students: \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\) Final Answer: 720 different ways
Q19: A school council has 12 members. In how many ways can a president, vice president, and secretary be chosen from the council?
Solution:
Ans: This is a permutation problem since the positions are different (order matters): \(P(12,3) = \frac{12!}{(12-3)!} = \frac{12!}{9!}\) \(= 12 \times 11 \times 10 = 1320\) Final Answer: 1,320 ways
Q20: A basketball coach must select 5 players from a team of 9 players to start the game. How many different starting lineups are possible?
Solution:
Ans: This is a combination problem since positions are not specified: \(C(9,5) = \frac{9!}{5!(9-5)!} = \frac{9!}{5! \times 4!}\) \(= \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}\) \(= \frac{3024}{24} = 126\) Final Answer: 126 different starting lineups
The document Worksheet (with Solutions): Counting, Permutations, and Combinations is a part of the Grade 9 Course Statistics & Probability.
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