Permutations and Combinations Important Questions.

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Practice these Permutations and Combinations questions for Class 11 Mathematics, each with the correct answer and a step-by-step explanation. Sign up free to practice all 27+ questions with adaptive difficulty.

1. Q1Hard

   What is the value of n such that P(n, 4) = 42 * P(n-1, 3), where n > 4?

   • A.7
   • B.8
   • C.10
   • D.12
   Answer: 10

   Explanation: The given equation represents a scenario where permutations of size 4 are equated to 42 times permutations of size 3. By solving the resulting quadratic equation for n, we find n = 10 is the valid solution. This involves breaking down the permutations into factorial expressions and simplifying step-by-step to determine the correct value of n.

2. Q2Hard

   Which of these expressions accurately represents the number of ways to select 3 items from a set of 5 when the order does not matter?

   • A.P(5,3)
   • B.r! × C(5,3)
   • C.C(5,3)
   • D.3!
   Answer: C(5,3)

   Explanation: The number of combinations, denoted by C(n,r), represents unordered selections. Here, since the order does not matter and we are selecting 3 items from 5, the expression C(5,3) is appropriate. It simplifies to finding the number of ways to choose a subset of size 3 from 5 distinct elements.

3. Q3Hard

   If you are arranging 4 distinct letters, and one letter is repeated twice, how many unique permutations exist?

   • A.4!
   • B.4! × 2!
   • C.4! / 2!
   • D.2! × 3!
   Answer: 4! / 2!

   Explanation: When dealing with repeated elements, you adjust the total permutations by dividing by the factorial of the number of repeated items. Here, with 4 distinct letters and one repeated twice, the unique permutations would be 4! divided by 2!, accounting for the indistinct arrangements of the repeated letter.

4. Q4Hard

   Based on historical context, which mathematician first provided general formulae for permutations and combinations?

   • A.Pingala
   • B.Sushruta
   • C.Mahavira
   • D.Jacob Bernoulli
   Answer: Mahavira

   Explanation: Historical records show that Mahavira, an ancient Jain mathematician around 850, was credited with providing general formulas for permutations and combinations. This knowledge was a substantial contribution to early combinatorial mathematics.

5. Q5Hard

   What is the correct relationship between permutations and combinations for n items taken r at a time?

   • A.P(n,r) = C(n,r)
   • B.P(n,r) = C(n,r) + r!
   • C.C(n,r) = r! × P(n,r)
   • D.P(n,r) = C(n,r) × r!
   Answer: P(n,r) = C(n,r) × r!

   Explanation: Each combination of r items from n can be arranged in r! ways. Hence, the total number of permutations (order matters) is given by multiplying the number of combinations by r factorials, showing the direct relationship between these two combinatorial concepts.

6. Q6Hard

   What is the value of 5P3 when computed using combinations?

   • A.10
   • B.30
   • C.40
   • D.60
   Answer: 60

   Explanation: By using the relationship P(n,r) = C(n,r) × r!, we find P(5,3) = C(5,3) × 3!. C(5,3) is 10, and 10 × 3! is computed to be 60, giving the correct number of ordered arrangements.

7. Q7Hard

   Which formula is used to calculate permutations where each event is independent and repetition is allowed?

   • A.n!
   • B.nCr
   • C.nPr
   • D.n^r
   Answer: n^r

   Explanation: Permutations with repetition involve independent choices for each position in the arrangement, allowing each element to be repeated. This scenario is represented by the formula n^r, where n is the number of choices and r is the number of positions.

8. Q8Hard

   How many different 7-digit numbers can be formed using the digits 1, 2, 4, 4, 0, 2, 2, with each digit used exactly once?

   • A.7!
   • B.7! × 2!
   • C.7! / (2! × 2!)
   • D.5! × 3!
   Answer: 7! / (2! × 2!)

   Explanation: When forming numbers with repeated digits, adjust for the identical digits by dividing by the factorial of their repetitions. Here, the two 2s and two 4s are repetitions, so the formula for unique permutations would be 7! divided by (2! × 2!).

9. Q9Hard

   Why is n! / (n - r)! the formula for permutations without repetition?

   • A.It ensures uniqueness across all items.
   • B.It accounts for the sequence of arrangements without replacing elements.
   • C.It divides by total permutations of the remaining items.
   • D.It is directly derived from nCr × r!
   Answer: It accounts for the sequence of arrangements without replacing elements.

   Explanation: The formula n! / (n - r)! calculates the number of ways to arrange r distinct items out of n, ensuring that each arrangement is counted exactly once. Factorial n! represents the total permutations, and dividing by (n - r)! corrects over-counting due to indistinct orderings of unused elements.

10. Q10Hard

    If you have 6 different items and you want to choose a subset of 3 items where the order doesn't matter, what formula do you use?

    • A.P(3,6)
    • B.C(3,6)
    • C.6C3
    • D.3P6
    Answer: 6C3

    Explanation: Combinations are used when order does not matter. The formula C(n,r), also written as 6C3, calculates the number of ways to select and arrange 3 items from 6 without regard to their specific order.

11. Q11Hard

    In a seating arrangement of 6 friends, how many ways can 3 friends be seated in 3 specific seats if order matters?

    • A.6C3
    • B.6! / 3!
    • C.C(3,6)
    • D.P(6,3)
    Answer: P(6,3)

    Explanation: For ordered seating arrangements, permutations are used. P(n,r) = P(6,3) calculates all possible ways to assign 3 of the 6 friends to 3 specific seats, ensuring that each friend is distinguished by their seat position.

12. Q12Hard

    Which of these expressions correctly represents the multiplication principle when three events occur in succession?

    • A.m + n + p
    • B.m × n / p
    • C.m + n × p
    • D.m × n × p
    Answer: m × n × p

    Explanation: The multiplication principle states that if one event can occur in m ways, a second in n ways, and a third in p ways, the total outcomes for the sequence is m × n × p. This captures all possible combinations through sequential independence.

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