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Quizzes > High School Quizzes > Electives

Example Combination Practice Quiz

Practice combination questions to master exam skills

Difficulty: Moderate
Grade: Grade 8
Study OutcomesCheat Sheet
Colorful paper art promoting Combo Conundrum, a math trivia for high school students.

Which of the following is an example of a combination?
Selecting 3 students from a group of 10 to form a club
Determining the finishing order of 3 runners
Arranging 3 students in a row from 10 for a photograph
Assigning roles of president, vice-president, and secretary to 3 people
This is a combination problem because the order of selection does not matter. Only the group chosen is important, not the arrangement of its members.
What is the correct formula to calculate the number of combinations when selecting r items out of n distinct items?
nCr = n! / (r!(n-r)!)
nCr = n! / (n*r!)
nPr = n! / (n-r)!
nCr = r! / (n!(n-r)!)
The correct formula for combinations is nCr = n! / (r!(n-r)!), which calculates the number of ways to select r items from n without considering order. This distinguishes combinations from permutations where order matters.
Which scenario illustrates the concept of combinations?
Assigning 4 different tasks to 4 employees
Arranging 4 books on a shelf
Selecting 4 students from a class of 25 for a project group
Ranking 4 runners by finish position
In this scenario, the order in which the 4 students are selected does not matter, making it a combination problem. The focus is solely on which students are chosen.
Which statement best describes combinations?
It uses subtraction instead of factorial in the formula.
Order does not matter; only the selection is important.
Order matters and every arrangement is unique.
Repetition of items is allowed.
Combinations are defined by the fact that order does not matter; only the items that are chosen are significant. This is the key difference between combinations and permutations.
How many ways can you choose 2 toppings from 4 available pizza toppings?
4
6
8
12
Using the formula for combinations, 4C2 = 4!/(2!2!) equals 6. This represents the number of different pairs of toppings that can be selected.
Calculate the number of ways to choose 3 members out of a 7-member group.
21
42
35
49
The number of combinations is calculated by 7C3 = 7!/(3!4!) which equals 35. This indicates there are 35 different ways to choose 3 members from 7.
How many combinations are possible when selecting 4 cards from a standard deck of 52 cards?
270000
635376
2598960
270725
The number of ways to choose 4 cards from 52 is given by the combination formula 52C4, which equals 270725. This counts all unordered selections of 4 cards from the deck.
If you have 10 different books, how many ways can you select 2 books to take on a trip?
20
120
90
45
Using the combination formula 10C2, the calculation is 10!/(2!8!) which equals 45. This represents the number of pairs of books you can choose.
Which of the following problems represents a combination rather than a permutation?
Scheduling 5 consecutive presentations
Arranging 5 students in a line
Ranking 5 contestants in a contest
Choosing 5 committee members from 12 candidates
Choosing committee members is a combination problem because the order in which they are chosen does not matter. Permutations, in contrast, require a specific order.
Calculate the value of 8C3.
512
336
24
56
The value of 8C3 is calculated as 8!/(3!5!) which equals 56. This represents the number of ways to choose 3 items from 8.
Find the number of ways to choose 0 items from any set of n items.
n
0
1
n-1
By definition, there is exactly one way to choose nothing from a set, which is the empty set. This is why nC0 is always equal to 1 regardless of the value of n.
In how many ways can a committee of 4 be formed from 9 people?
100
126
84
210
The number of combinations for forming a committee of 4 from 9 people is given by 9C4 = 9!/(4!5!) which equals 126. This counts all the possible groups of 4 regardless of order.
How does increasing the total number of items, n, affect the number of combinations when the number selected, r, is fixed?
It makes the outcome unpredictable.
It has no effect on the number of combinations.
It always decreases the number of combinations.
It increases the number of combinations.
When n increases while r remains fixed, there are more items available to choose from, thereby increasing the total number of combinations. This is a direct consequence of the combination formula.
What is the value of 12C6?
924
1320
792
1001
Calculating 12C6 using the formula 12!/(6!6!) results in 924. This is the number of ways to choose 6 items from 12 without regard to the order.
Which of the following statements about 10C3 is correct?
10C3 equals 720.
10C3 equals 100.
10C3 equals 120.
10C3 equals 30.
Using the combination formula, 10C3 is calculated as 10!/(3!7!) which equals 120. This confirms there are 120 ways to choose 3 items from 10 when order is not considered.
For a fixed set of n items, how does the number of combinations change as you increase the number of items chosen from 0 to n?
It remains constant regardless of r.
It increases continuously and then remains constant.
It increases until reaching a maximum at r = n/2 and then decreases symmetrically.
It decreases continuously.
Combination values increase as r increases from 0 until reaching a maximum near r = n/2, after which they decrease. This symmetric behavior is due to the identity nCr = nC(n-r).
How many ways can you choose 5 elements from a set of 20, and what symmetry property does this number demonstrate?
15504; this demonstrates that 20C5 equals 20C15.
15504; this shows that the combinations increase linearly with n.
38760; this demonstrates the inverse relationship of combinations.
4845; this shows that combinations are always prime numbers.
Calculating 20C5 gives 15504, and by the symmetry property nCr = nC(n-r), it equals 20C15. This demonstrates the inherent symmetry in combination calculations.
A sports team has 12 players. In how many ways can 5 players be chosen for a drill if the order of selection does not matter?
1000
950
480
792
The number of ways to choose 5 players from 12 is found using the combination formula 12C5, which equals 792. This calculation ignores the order of selection, as required in combination problems.
Determine the number of combinations possible when picking 3 out of 3 items, and what does this imply?
0; because choosing all items is not allowed.
None; the result is undefined.
1; there is only one way to choose all items.
3; there are three ways because each item can be chosen separately.
When all items in a set are chosen, there is exactly one way to do so, as represented by the identity nCn = 1. This underlines a fundamental property of combinations.
Two groups are forming separate committees: one group of 8 students forms a committee of 3, and another group of 6 students forms a committee of 2. How many total ways can both committees be formed?
900
960
750
840
The total number of ways to form both committees is the product of the individual combinations: 8C3 (56 ways) multiplied by 6C2 (15 ways), which equals 840. This application of the multiplication principle combines independent selection processes.
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Study Outcomes

  1. Identify and differentiate between combinations and permutations in problem-solving contexts.
  2. Apply counting principles to solve combination problems effectively.
  3. Analyze various examples to determine when a combination approach is appropriate.
  4. Evaluate problem statements to extract key mathematical concepts and apply relevant strategies.
  5. Synthesize learned concepts from different math topics to enhance exam readiness.

Combination Example Cheat Sheet

  1. Combinations ignore order - In combinations, the sequence of picking items doesn't matter. Whether you choose Alice → Bob → Charlie or Charlie → Bob → Alice, it's counted as one group! GeeksforGeeks
  2. The combination formula - Remember C(n, r) = n! / [r! × (n - r)!], where n is the total pool and r is your pick size. This neat formula spits out exactly how many unique groups you can form without worrying about order. Wumbo Formulas
  3. Permutations vs. combinations - Permutations care about order; combinations do not. Lining up books in a row is a permutation, but grabbing books off the shelf for a reading list is pure combination action! GeeksforGeeks
  4. Practice makes perfect - Tackle committee-building or lottery-style problems to cement your combo skills. The more real questions you crunch, the more naturally the formula and concepts will click. Statistics by Jim
  5. Everyday combo examples - Picking pizza toppings is a classic combo scenario - pepperoni, mushrooms, and olives taste the same regardless of the order you sprinkle them. Spotting these real-life cases makes the math feel less abstract. Statistics by Jim
  6. Factorials are your friends - Factorial notation (n!) multiplies all whole numbers from 1 up to n. Knowing how to break down 5! = 5×4×3×2×1 will save tons of time when plugging into the combo formula. Wumbo Formulas
  7. Apply it everywhere - From team selection to playlist shuffles, try using C(n, r) in daily puzzles. Real-world applications help you see patterns and understand why combinations rule group selections. KeyDifferences
  8. Probability power-up - Combinations are the backbone of many probability and statistics problems - think lottery odds or card draws. Mastering combos means you can tackle event-likelihood questions like a pro. ThoughtCo
  9. Scenario spotting - Before solving, ask: "Does order matter here?" If not, you're in combination territory. Practicing this quick check will boost your problem-solving speed and confidence. Statistics by Jim
  10. Mental mnemonic - Use "C for Combination, C for Committee" to recall that combos pick groups without order. A catchy phrase like this can turn tricky questions into memory-friendly puzzles. Socratic
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