Q1: In how many ways can 5 different books be arranged on a shelf? (a) 25 (b) 120 (c) 60 (d) 100
Solution:
Ans: (b) Explanation: The number of ways to arrange \(n\) distinct objects is \(n!\). For 5 books, this is \(5! = 5 × 4 × 3 × 2 × 1 = 120\). Option (a) represents \(5^2\), option (c) is \(5 × 12\), and option (d) is arbitrary, all of which are common errors in permutation calculations.
Q2: How many different 3-letter "words" (arrangements) can be formed using the letters A, B, C, D, E without repetition? (a) 15 (b) 125 (c) 60 (d) 20
Solution:
Ans: (c) Explanation: This is a permutation problem: \(P(5,3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{120}{2} = 60\). Option (a) represents combinations \(C(5,3) = 10\) with an error, option (b) assumes repetition allowed \(5^3 = 125\), and option (d) is \(5 × 4\) missing the third selection.
Q3: A committee of 4 students is to be selected from a group of 10 students. How many different committees can be formed? (a) 5040 (b) 210 (c) 40 (d) 252
Solution:
Ans: (b) Explanation: Since order does not matter in selecting a committee, we use combinations: \(C(10,4) = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = \frac{10 × 9 × 8 × 7}{4 × 3 × 2 × 1} = \frac{5040}{24} = 210\). Option (a) is \(10 × 9 × 8 × 7\) (permutation error), option (c) is \(10 × 4\), and option (d) is \(C(10,5)\).
Q4: A fair six-sided die is rolled once. What is the probability of rolling a number greater than 4? (a) \(\frac{1}{6}\) (b) \(\frac{1}{3}\) (c) \(\frac{1}{2}\) (d) \(\frac{2}{3}\)
Solution:
Ans: (b) Explanation: The favorable outcomes are 5 and 6, giving 2 favorable outcomes out of 6 total outcomes. Thus, \(P(\text{greater than 4}) = \frac{2}{6} = \frac{1}{3}\). Option (a) represents one outcome, option (c) represents three outcomes, and option (d) represents four outcomes, all common counting errors.
Q5: In how many ways can the letters of the word "BOOK" be arranged? (a) 24 (b) 12 (c) 6 (d) 4
Solution:
Ans: (b) Explanation: The word "BOOK" has 4 letters with the letter O repeated twice. The number of arrangements is \(\frac{4!}{2!} = \frac{24}{2} = 12\). Option (a) assumes all letters are distinct (\(4!\)), option (c) might confuse with \(3!\), and option (d) is simply the number of letters, representing common errors with repeated elements.
Q6: A bag contains 3 red balls and 5 blue balls. If one ball is drawn at random, what is the probability it is red? (a) \(\frac{3}{5}\) (b) \(\frac{5}{8}\) (c) \(\frac{3}{8}\) (d) \(\frac{1}{3}\)
Solution:
Ans: (c) Explanation: The total number of balls is \(3 + 5 = 8\). The probability of drawing a red ball is \(\frac{3}{8}\). Option (a) compares red to blue balls only, option (b) is the probability of blue, and option (d) assumes only 9 balls or other miscounting.
Q7: How many different 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if repetition is not allowed? (a) 120 (b) 625 (c) 20 (d) 24
Solution:
Ans: (a) Explanation: This is a permutation: \(P(5,4) = \frac{5!}{(5-4)!} = \frac{5!}{1!} = 120\). Option (b) represents \(5^4\) with repetition allowed, option (c) is \(5 × 4\), and option (d) is \(4!\), all representing common computational mistakes.
Q8: Two dice are rolled simultaneously. What is the probability that the sum of the numbers is 7? (a) \(\frac{1}{6}\) (b) \(\frac{1}{12}\) (c) \(\frac{5}{36}\) (d) \(\frac{7}{36}\)
Solution:
Ans: (a) Explanation: The total number of outcomes when rolling two dice is \(6 × 6 = 36\). The favorable outcomes for a sum of 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), which is 6 outcomes. Thus, \(P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}\). Option (b) divides by 12 instead of 36, option (c) counts 5 outcomes, and option (d) counts 7 outcomes, representing common counting errors.
Section B: Fill in the Blanks
Q9: The number of ways to arrange \(n\) distinct objects in a row is __________.
Solution:
Ans: \(n!\) Explanation: The factorial function \(n!\) represents the number of permutations of \(n\) distinct objects, calculated as \(n × (n-1) × (n-2) × ... × 2 × 1\).
Q10: The formula for the number of combinations of \(n\) objects taken \(r\) at a time is __________.
Solution:
Ans: \(C(n,r) = \frac{n!}{r!(n-r)!}\) Explanation: This is the combination formula, used when the order of selection does not matter. It divides the permutation count by \(r!\) to eliminate ordering.
Q11: If an event is certain to occur, its probability is __________.
Solution:
Ans: 1 Explanation: A certain event has a probability of 1, meaning it will definitely happen. Probabilities range from 0 (impossible) to 1 (certain).
Q12: The sum of the probabilities of all possible outcomes in a sample space is __________.
Solution:
Ans: 1 Explanation: This is a fundamental property of probability. Since the sample space includes all possible outcomes, their probabilities must sum to 1.
Q13: The number of permutations of \(n\) objects taken \(r\) at a time is denoted by __________ and equals \(\frac{n!}{(n-r)!}\).
Solution:
Ans: \(P(n,r)\) or \(_nP_r\) Explanation: The notation \(P(n,r)\) or \(_nP_r\) represents permutations, where order matters. The formula accounts for selecting and arranging \(r\) objects from \(n\) total objects.
Q14: If two events A and B are mutually exclusive, then \(P(A \cap B) = __________\).
Solution:
Ans: 0 Explanation:Mutually exclusive events cannot occur simultaneously, so the probability of their intersection is 0. This means there is no overlap between the two events.
Section C: Word Problems
Q15: A pizza shop offers 3 types of crust, 4 types of sauce, and 6 types of toppings. If a customer must choose exactly one of each, how many different pizza combinations are possible?
Solution:
Ans: Step 1: Use the fundamental counting principle. Step 2: Total combinations = \(3 × 4 × 6 = 72\) Final Answer: 72 different pizza combinations
Q16: A student council has 8 members. In how many ways can a president, vice president, and secretary be chosen if one person cannot hold more than one position?
Solution:
Ans: Step 1: This is a permutation problem since the positions are distinct and order matters. Step 2: Number of ways = \(P(8,3) = \frac{8!}{(8-3)!} = \frac{8!}{5!} = 8 × 7 × 6 = 336\) Final Answer: 336 ways
Q17: A standard deck of 52 cards contains 4 aces. If one card is drawn at random from the deck, what is the probability that it is an ace?
Solution:
Ans: Step 1: Number of favorable outcomes = 4 (aces) Step 2: Total number of outcomes = 52 (cards) Step 3: Probability = \(\frac{4}{52} = \frac{1}{13}\) Final Answer: \(\frac{1}{13}\)
Q18: A basketball team has 12 players. The coach needs to select 5 players to be the starting lineup. How many different starting lineups can be formed?
Solution:
Ans: Step 1: Since the order does not matter for a lineup selection, use combinations. Step 2: \(C(12,5) = \frac{12!}{5!(12-5)!} = \frac{12!}{5!7!}\) Step 3: \(= \frac{12 × 11 × 10 × 9 × 8}{5 × 4 × 3 × 2 × 1} = \frac{95040}{120} = 792\) Final Answer: 792 different starting lineups
Q19: A bag contains 4 green marbles, 5 yellow marbles, and 6 purple marbles. If two marbles are drawn at random without replacement, what is the probability that both marbles are green?
Solution:
Ans: Step 1: Total marbles = \(4 + 5 + 6 = 15\) Step 2: Probability first marble is green = \(\frac{4}{15}\) Step 3: After drawing one green marble, 3 green marbles remain out of 14 total. Step 4: Probability second marble is green = \(\frac{3}{14}\) Step 5: Probability both are green = \(\frac{4}{15} × \frac{3}{14} = \frac{12}{210} = \frac{2}{35}\) Final Answer: \(\frac{2}{35}\)
Q20: How many different ways can 7 students be seated in a row if 2 particular students must sit next to each other?
Solution:
Ans: Step 1: Treat the 2 particular students as a single unit. Step 2: Now we have 6 units to arrange (the pair + 5 other students): \(6! = 720\) Step 3: The 2 students within their unit can be arranged in \(2! = 2\) ways. Step 4: Total arrangements = \(6! × 2! = 720 × 2 = 1440\) Final Answer: 1440 ways
The document Worksheet (with Solutions): Combinatorics and Probability is a part of the Grade 9 Course Statistics & Probability.
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