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Mathematics Derivations of formulas and their connections MCQ Questions for Class 11 General

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Question 121/600 Easy Mathematics Permutations and Combinations Derivations of formulas and their connections Class 11 Level 66

\(^{n}P_r\) के recurrence (^{n}P_r=^{n}P_{r-1}(n-r+1)) में नया factor क्या दर्शाता है?

What does the new factor represent in the recurrence (^{n}P_r=^{n}P_{r-1}(n-r+1))?

Explanation opens after your attempt
Correct Answer

A. (r)वें स्थान के विकल्पChoices for the (r)th position

Explanation

Simple Explanation

पहले (r-1) स्थान भरने के बाद (r)वें स्थान पर (n-r+1) विकल्प बचते हैं। परीक्षा में recurrence को next-place idea से जोड़ें। / After filling the first (r-1) positions (n-r+1) choices remain for the (r)th position. In exams connect recurrence with the next-place idea.

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Question 122/600 Easy Mathematics Permutations and Combinations Derivations of formulas and their connections Class 11 Level 66

\(^{n}C_r\) के ratio \(\frac{^{n}C_{r+1}}{^{n}C_r}\) का सही मान कौन-सा है?

What is the correct value of the ratio \(\frac{^{n}C_{r+1}}{^{n}C_r}\)?

Explanation opens after your attempt
Correct Answer

B. \(\frac{n-r}{r+1}\)

Explanation

Simple Explanation

Factorial formula को divide करने पर \(\frac{n-r}{r+1}\) मिलता है। परीक्षा में adjacent combinations के लिए ratio से तेजी आती है। / Dividing the factorial formulas gives \(\frac{n-r}{r+1}\). In exams ratios make adjacent combinations faster.

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Question 123/600 Easy Mathematics Permutations and Combinations Derivations of formulas and their connections Class 11 Level 66

यदि \(^{n}C_r=^{n}C_{r+2}\) हो तो कौन-सा relation संभव है?

If \(^{n}C_r=^{n}C_{r+2}\) which relation is possible?

Explanation opens after your attempt
Correct Answer

A. (r+r+2=n)

Explanation

Simple Explanation

Equal combinations में lower indices समान या पूरक होते हैं। यहाँ अलग indices हैं इसलिए (r+(r+2)=n) होगा। / In equal combinations the lower indices are equal or complementary. Here the indices are different so (r+(r+2)=n).

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Question 124/600 Easy Mathematics Permutations and Combinations Derivations of formulas and their connections Class 11 Level 66

\(^{14}C_5=^{14}C_s\) और \(s\neq5\) हो तो (s) क्या होगा?

If \(^{14}C_5=^{14}C_s\) and \(s\neq5\) what is (s)?

Explanation opens after your attempt
Correct Answer

C. (9)

Explanation

Simple Explanation

पूरक index (14-5=9) है। परीक्षा में equal combination में दूसरा index (n-r) देखें। / The complementary index is (14-5=9). In exams look for (n-r) as the second index in equal combinations.

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Question 125/600 Easy Mathematics Permutations and Combinations Derivations of formulas and their connections Class 11 Level 66

\(^{n}C_n=1\) को पूरक identity से किसके बराबर समझ सकते हैं?

Using complementary identity \(^{n}C_n=1\) can be understood as equal to what?

Explanation opens after your attempt
Correct Answer

B. \(^{n}C_0\)

Explanation

Simple Explanation

\(^{n}C_n=^{n}C_0\) और दोनों का मान (1) है। परीक्षा में all selected और none selected दोनों boundary cases याद रखें। / \(^{n}C_n=^{n}C_0\) and both have value (1). In exams remember both all selected and none selected boundary cases.

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Question 126/600 Easy Mathematics Permutations and Combinations Derivations of formulas and their connections Class 11 Level 66

यदि (n) अलग वस्तुओं में से कोई एक वस्तु खास मानकर चुनी जाए तो \(^{n}C_r\) की included case count क्या होगी?

If one object is treated as special among (n) distinct objects what is the included case count for \(^{n}C_r\)?

Explanation opens after your attempt
Correct Answer

B. \(^{n-1}C_{r-1}\)

Explanation

Simple Explanation

खास वस्तु चुन ली गई तो बाकी (r-1) वस्तुएँ (n-1) में से चुननी हैं। परीक्षा में included case में lower index (1) घटता है। / If the special object is selected the remaining (r-1) objects are chosen from (n-1). In exams the lower index decreases by (1) in the included case.

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Question 127/600 Easy Mathematics Permutations and Combinations Derivations of formulas and their connections Class 11 Level 66

यदि खास वस्तु \(^{n}C_r\) के selection में शामिल न हो तो count क्या होगी?

If the special object is not included in the selection for \(^{n}C_r\) what is the count?

Explanation opens after your attempt
Correct Answer

A. \(^{n-1}C_r\)

Explanation

Simple Explanation

खास वस्तु हटाने के बाद (n-1) वस्तुओं में से पूरे (r) चुनने हैं। परीक्षा में excluded case में lower index वही रहता है। / After removing the special object all (r) objects are chosen from (n-1). In exams the lower index remains the same in the excluded case.

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Question 128/600 Easy Mathematics Permutations and Combinations Derivations of formulas and their connections Class 11 Level 66

\(^{10}C_4\) को एक fixed object method से लिखने पर कौन-सा decomposition सही है?

Using the fixed object method which decomposition of \(^{10}C_4\) is correct?

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Correct Answer

A. \(^{9}C_4+^{9}C_3\)

Explanation

Simple Explanation

एक fixed object को include या exclude करने से \(^{9}C_3\) और \(^{9}C_4\) मिलते हैं। परीक्षा में Pascal split में upper index (1) घटता है। / Including or excluding one fixed object gives \(^{9}C_3\) and \(^{9}C_4\). In exams the upper index decreases by (1) in a Pascal split.

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Question 129/600 Easy Mathematics Permutations and Combinations Derivations of formulas and their connections Class 11 Level 66

(4) अलग अक्षरों से repetition allowed होने पर (3)-letter codes की संख्या क्या होगी?

How many (3)-letter codes can be made from (4) distinct letters when repetition is allowed?

Explanation opens after your attempt
Correct Answer

C. \(4^3\)

Explanation

Simple Explanation

हर स्थान पर (4) विकल्प बार-बार उपलब्ध हैं। परीक्षा में repetition allowed हो तो power formula \(n^r\) लगाएँ। / Each position has (4) choices again and again. In exams use the power formula \(n^r\) when repetition is allowed.

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Question 130/600 Easy Mathematics Permutations and Combinations Derivations of formulas and their connections Class 11 Level 66

(4) अलग अक्षरों से repetition not allowed होने पर (3)-letter codes की संख्या क्या होगी?

How many (3)-letter codes can be made from (4) distinct letters when repetition is not allowed?

Explanation opens after your attempt
Correct Answer

B. \(^{4}P_3\)

Explanation

Simple Explanation

बिना पुनरावृत्ति ordered code के लिए \(4\times3\times2\) तरीके हैं। परीक्षा में code में order होता है। / For an ordered code without repetition there are \(4\times3\times2\) ways. In exams a code has order.

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