Class 11 Mathematics Permutations and Combinations Derivations of formulas and their connections Hard Level 65

(n) objects में से (r) objects चुनने में exactly (s) special objects लेने का सूत्र क्या है?

What is the formula for choosing (r) objects from (n) objects with exactly (s) special objects?

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Correct Answer

A. \(^{a}C_s\cdot{}^{n-a}C_{r-s}\) जहां special objects (a) हैं\(^{a}C_s\cdot{}^{n-a}C_{r-s}\) where there are (a) special objects

Step 1

Concept

Choose (s) objects from the special group and (r-s) from the non-special group. In exams split exactly conditions into product of choices.

Step 2

Why this answer is correct

The correct answer is A. \(^{a}C_s\cdot{}^{n-a}C_{r-s}\) जहां special objects (a) हैं / \(^{a}C_s\cdot{}^{n-a}C_{r-s}\) where there are (a) special objects. Choose (s) objects from the special group and (r-s) from the non-special group. In exams split exactly conditions into product of choices.

Step 3

Exam Tip

Special group से (s) और non-special group से (r-s) objects चुनते हैं। परीक्षा में exactly condition को product of choices में तोड़ें।

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(n) objects में से (r) objects चुनने में exactly (s) special objects लेने का सूत्र क्या है? / What is the formula for choosing (r) objects from (n) objects with exactly (s) special objects?

Correct Answer: A. \(^{a}C_s\cdot{}^{n-a}C_{r-s}\) जहां special objects (a) हैं / \(^{a}C_s\cdot{}^{n-a}C_{r-s}\) where there are (a) special objects. Explanation: Special group से (s) और non-special group से (r-s) objects चुनते हैं। परीक्षा में exactly condition को product of choices में तोड़ें। / Choose (s) objects from the special group and (r-s) from the non-special group. In exams split exactly conditions into product of choices.

Which concept should I revise for this Mathematics MCQ?

Choose (s) objects from the special group and (r-s) from the non-special group. In exams split exactly conditions into product of choices.

What exam hint can help solve this Mathematics question?

Special group से (s) और non-special group से (r-s) objects चुनते हैं। परीक्षा में exactly condition को product of choices में तोड़ें।