JEE Main Algebra — Combinatorics & Series: Permutations and Combinations – Chapter Test

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Difficulty: 10 medium · 10 hard

Permutations and Combinations MCQ questions with answers

The first 3 of 20 questions. Unlock the series for the rest, each with a solution.

1. Q1. The number of ways in which six '+' signs and four '−' signs can be arranged in a row so that no two '−' signs occur together is:

   छह '+' चिह्नों और चार '−' चिह्नों को एक पंक्ति में इस प्रकार व्यवस्थित करने के तरीकों की संख्या कि कोई भी दो '−' चिह्न एक साथ न आएँ, है:

   • A. 15
   • B. 35
   • C. 7
   • D. 210
   Solution

   Place the six '+' signs in a row; they create 7 gaps (including the ends). Choosing 4 of these gaps for the '−' signs gives ⁷C₄ = 35 arrangements.

   छह '+' चिह्नों को एक पंक्ति में रखिए; वे 7 रिक्त स्थान (सिरों सहित) बनाते हैं। '−' चिह्नों के लिए इनमें से 4 रिक्त स्थान चुनने पर ⁷C₄ = 35 व्यवस्थाएँ प्राप्त होती हैं।

2. Q2. The value of the sum ⁴⁷C₄ + ⁵¹C₃ + ⁵⁰C₃ + ⁴⁹C₃ + ⁴⁸C₃ + ⁴⁷C₃ is equal to:

   योग ⁴⁷C₄ + ⁵¹C₃ + ⁵⁰C₃ + ⁴⁹C₃ + ⁴⁸C₃ + ⁴⁷C₃ का मान बराबर है:

   • A. ⁵³C₄
   • B. ⁵¹C₄
   • C. ⁵²C₄
   • D. ⁵²C₃
   Solution

   Using nCr + nC(r-1) = (n+1)Cr repeatedly: ⁴⁷C₄ + ⁴⁷C₃ = ⁴⁸C₄, then + ⁴⁸C₃ = ⁴⁹C₄, continuing up to + ⁵¹C₃ = ⁵²C₄.

   nCr + nC(r-1) = (n+1)Cr का बार-बार उपयोग करते हुए: ⁴⁷C₄ + ⁴⁷C₃ = ⁴⁸C₄, फिर + ⁴⁸C₃ = ⁴⁹C₄, इसी प्रकार + ⁵¹C₃ = ⁵²C₄ तक।

3. Q3. Consider the following statements: 1. ⁿCᵣ = ⁿCₙ₋ᵣ for all integers 0 ≤ r ≤ n. 2. ⁿPᵣ = r! × ⁿCᵣ. 3. ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ. Which of these statements is/are correct?

   निम्नलिखित कथनों पर विचार कीजिए: 1. सभी पूर्णांकों 0 ≤ r ≤ n के लिए ⁿCᵣ = ⁿCₙ₋ᵣ। 2. ⁿPᵣ = r! × ⁿCᵣ। 3. ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ। इनमें से कौन-सा/से कथन सही है/हैं?

   • A. 1 and 3 only / केवल 1 और 3
   • B. 2 and 3 only / केवल 2 और 3
   • C. 1 and 2 only / केवल 1 और 2
   • D. 1, 2 and 3 / 1, 2 और 3
   Solution

   All three are standard identities: selecting r objects is equivalent to rejecting n−r objects; every selection of r objects can be arranged among themselves in r! ways; and Pascal's rule gives ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ.

   तीनों मानक सर्वसमिकाएँ हैं: r वस्तुओं का चयन करना n−r वस्तुओं को छोड़ने के तुल्य है; r वस्तुओं के प्रत्येक चयन को आपस में r! प्रकार से व्यवस्थित किया जा सकता है; तथा पास्कल के नियम से ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ प्राप्त होता है।

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