Multiple Choice Questions
Discrete Mathematics: Combinatorics and Probability
Topic: Combinatorics
Grade: 11
Question 1:
In how many ways can a committee of 3 students be chosen from a class of 10 students?
a) 10
b) 30
c) 45
d) 120
Answer: c) 45
Explanation: The number of ways to choose a committee of 3 students from a class of 10 students can be calculated using the combination formula. The formula for combinations is nCr = n! / (r!(n-r)!), where n is the total number of students and r is the number of students to be chosen. In this case, n = 10 and r = 3, so the number of ways is 10C3 = 10! / (3!(10-3)!) = 120 / (6*7*8) = 45.
Example: Let\’s say the class has 10 students labeled A, B, C, D, E, F, G, H, I, and J. We can choose a committee of 3 students as follows: ABC, ABD, ABE, …, FGJ, FHI, FHI, GHI, GHI. There are a total of 45 possible combinations.
Question 2:
A group of 5 friends are going to a concert and they want to sit together in a row. If there are 10 seats available, how many different seating arrangements are possible?
a) 10
b) 50
c) 100
d) 120
Answer: d) 120
Explanation: The number of different seating arrangements can be calculated using the permutation formula. The formula for permutations is nPr = n! / (n-r)!, where n is the total number of seats and r is the number of friends in the group. In this case, n = 10 and r = 5, so the number of arrangements is 10P5 = 10! / (10-5)! = 10! / 5! = 10*9*8*7*6 = 1200 / 120 = 120.
Example: Let\’s say the seats are labeled 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. The friends can sit in the following arrangements: 12345, 23456, 34567, …, 678910. There are a total of 120 possible arrangements.
Question 3:
A committee of 4 people is to be chosen from a group of 8 men and 6 women. How many different committees can be formed if there must be at least one man and one woman in the committee?
a) 280
b) 336
c) 420
d) 504
Answer: b) 336
Explanation: To calculate the number of different committees, we need to consider the different combinations of men and women that can be chosen. We can choose 1 man and 3 women, 2 men and 2 women, 3 men and 1 woman, or 4 men and 0 women. The number of ways to choose 1 man and 3 women is 8C1 * 6C3 = 8 * 20 = 160. The number of ways to choose 2 men and 2 women is 8C2 * 6C2 = 28 * 15 = 420. The number of ways to choose 3 men and 1 woman is 8C3 * 6C1 = 56 * 6 = 336. The number of ways to choose 4 men and 0 women is 8C4 * 6C0 = 70 * 1 = 70. Adding up these possibilities gives a total of 160 + 420 + 336 + 70 = 986.
Example: Let\’s say the group of men is labeled A, B, C, D, E, F, G, and H, and the group of women is labeled W, X, Y, Z, U, and V. Possible committees could be AWXY, BCDU, EFVZ, GHYU, etc. There are a total of 336 possible committees.
Question 4:
A bag contains 5 red balls, 4 blue balls, and 3 green balls. If 3 balls are randomly drawn from the bag without replacement, what is the probability that all 3 balls are red?
a) 1/11
b) 5/12
c) 10/33
d) 5/99
Answer: c) 10/33
Explanation: The probability of drawing 3 red balls can be calculated by dividing the number of favorable outcomes (drawing 3 red balls) by the total number of possible outcomes (drawing any 3 balls). The number of favorable outcomes is given by 5C3 (choosing 3 red balls from the 5 available), and the total number of possible outcomes is given by 12C3 (choosing any 3 balls from the 12 available). So the probability is 5C3 / 12C3 = 10 / 220 = 10/33.
Example: Let\’s say the red balls are labeled R1, R2, R3, R4, and R5. Possible outcomes could be R1R2R3, R1R2R4, R1R2R5, R1R3R4, R1R3R5, R1R4R5, R2R3R4, R2R3R5, R2R4R5, R3R4R5. There are a total of 10 favorable outcomes out of 33 possible outcomes.
Question 5:
In a deck of 52 playing cards, how many 5-card hands contain exactly 3 aces?
a) 1680
b) 1287
c) 64
d) 128
Answer: d) 128
Explanation: To calculate the number of 5-card hands containing exactly 3 aces, we need to consider the number of ways to choose 3 aces from the 4 available and 2 cards from the remaining 48 non-ace cards. The number of ways to choose 3 aces is 4C3 = 4, and the number of ways to choose 2 cards from the remaining 48 is 48C2 = 1128. Multiplying these possibilities gives a total of 4 * 1128 = 4512. However, we need to divide by 5C3 (the number of ways to arrange the 3 aces within the 5-card hand) to get the final answer. So the number of 5-card hands containing exactly 3 aces is 4512 / 5C3 = 128.
Example: Let\’s say the aces are labeled Aâ™ , A♥, A♦, and A♣. Possible 5-card hands could be Aâ™ A♥A♦2♣, Aâ™ A♥A♦3♣, Aâ™ A♥A♦4♣, …, Aâ™ A♥A♦K♣. There are a total of 128 possible 5-card hands containing exactly 3 aces.