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Subjects

Mathematics

Derivations of formulas and their connections

सूत्रों की व्युत्पत्तियाँ और उनके पारस्परिक संबंध

In Class 11 Mathematics, this topic explains how the formulas in Permutations and Combinations are derived and how they are connected. Students learn the meaning of factorial notation, develop the formulas for nPr and nCr from counting principles, and understand why nPr = r! nCr and nCr = nC(n−r). The topic also shows when to use arrangements or selections, helping students apply these relationships logically instead of relying on memorized formulas.

TOPIC PRACTICE

Quiz this set

Up to 20 questions from this page. Select your focus, then start.

20 questions

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Hard · Level 6
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  1. (^{6}C_5\cdot5!)
  2. (^{7}C_5\cdot5!)
  3. (^{11}C_5)
  4. (5!)
Hard · Level 6
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  1. (5!\cdot5!)
  2. (2\cdot5!\cdot5!)
  3. (10!)
  4. (^{10}C_5)
Hard · Level 6
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  1. (7!)
  2. (\frac{7!}{2})
  3. (8!)
  4. (\frac{8!}{2})
Hard · Level 6
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  1. One person is fixed to remove rotations
  2. Two people are fixed to remove reflections
  3. All people are identical
  4. Order has no importance
Hard · Level 6
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  1. (5!\cdot2^5)
  2. (4!\cdot2^5)
  3. (10!)
  4. (\frac{9!}{2^5})
Hard · Level 6
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  1. Because (A) and (R) appear twice each
  2. Because (G) and (E) appear twice each
  3. Because the total letters are (7)
  4. Because vowels are together
Hard · Level 6
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  1. (3!3!2!)
  2. (2!2!2!)
  3. (10!)
  4. (4!3!)
Hard · Level 6
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  1. (\frac{8!}{3!})
  2. (8!\cdot3!)
  3. (5!\cdot3!)
  4. (^{8}C_3)
Hard · Level 6
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  1. (10!)
  2. (\frac{10!}{3!})
  3. (7!\cdot3!)
  4. (^{10}C_3)
Hard · Level 6
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  1. (0,2,4,6)
  2. (1,3,5)
  3. (0,1,2,3,4,5,6)
  4. (2,4,6)
Hard · Level 6
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  1. (6^3)
  2. (5\cdot6\cdot6)
  3. (^{6}P_3)
  4. (^{6}C_3)
Hard · Level 6
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  1. Each position has (7) independent choices
  2. Choices decrease at each position
  3. Order is ignored
  4. All symbols are identical
Hard · Level 6
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  1. (7^5)
  2. (^{7}C_5)
  3. (^{7}P_5)
  4. (^{12}C_5)
Hard · Level 6
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  1. Because (b) is chosen from (r) brackets
  2. Because (a) is chosen from (r) brackets
  3. Because all terms are arranged
  4. Because the coefficient is (r!)
Hard · Level 6
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  1. Adding all possible subset sizes
  2. Counting only (r)-sized subsets
  3. Arranging all objects
  4. Dividing every subset by (n!)
Hard · Level 6
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  1. Choosing r objects from n objects can be counted by choosing the \(n-r\) objects left unchosen.
  2. Every change in the order of r selected objects creates a new combination.
  3. Choosing r objects from n is the same as choosing r objects with repetition.
  4. \(^{n}C_r\) equals \(^{n}C_{n-r}\) only when n is even.
Hard · Level 6
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  1. Marking one selected member
  2. Marking two rejected members
  3. Arranging all members in a circle
  4. Dividing every subset by (r!)
Hard · Level 6
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  1. (^{n}C_2 2^{n-2})
  2. (^{n}C_2 2^n)
  3. (n2^{n-1})
  4. (2^n)
Hard · Level 6
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  1. Pascal identity
  2. Vandermonde identity
  3. Circular identity
  4. Factorial identity only
Hard · Level 6
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  1. (\frac{n!}{a!b!(n-a-b)!})
  2. (\frac{n!}{(a+b)!})
  3. ({}^{n}P_a{}^{n}P_b)
  4. (n^{a+b})

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