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

Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.

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Derivations of formulas and their connections Practice Questions

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(10) distinct beads की necklace arrangements में rotations same लेकिन reflections different हों, तो count क्या है?

For necklace arrangements of (10) distinct beads where rotations are the same but reflections are different, what is the count?

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Correct Answer

A. (9!)

Explanation

Simple Explanation

केवल rotations duplicate हैं इसलिए circular count ((10-1)!) है। परीक्षा में reflection condition पढ़कर ही (2) से divide करें। / Only rotations are duplicates, so the circular count is ((10-1)!). In exams divide by (2) only after reading the reflection condition.

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(10) distinct beads की bracelet arrangements में count क्या होगा?

What is the count for bracelet arrangements of (10) distinct beads?

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Correct Answer

A. \(\frac{9!}{2}\)

Explanation

Simple Explanation

Bracelet में rotations और reflections दोनों same मानी जाती हैं। परीक्षा में bracelet के लिए (\frac{(n-1)!}{2}) use करें। / In a bracelet, both rotations and reflections are considered the same. In exams use (\frac{(n-1)!}{2}) for bracelets.

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Digits (0,1,2,3,4,5,6,7,8,9) से repetition बिना (6)-digit numbers बनते हैं और number even हो। (0) last digit case का count क्या है?

Using digits (0,1,2,3,4,5,6,7,8,9) without repetition, (6)-digit even numbers are formed. What is the count when (0) is the last digit?

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Correct Answer

A. \(^{9}P_5\)

Explanation

Simple Explanation

Last digit (0) fix होने पर first place पर zero issue नहीं रहता और (9) non-zero digits से (5) places भरते हैं। परीक्षा में zero-last case अलग करें। / When the last digit is fixed as (0), there is no leading-zero issue and (5) places are filled from (9) non-zero digits. In exams separate the zero-last case.

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Digits (0) से (9) तक repetition बिना (6)-digit even numbers में non-zero even last digit case का count क्या होगा?

Using digits (0) to (9) without repetition, what is the count for (6)-digit even numbers with a non-zero even last digit?

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Correct Answer

A. \(4\cdot8\cdot{}^{8}P_4\)

Explanation

Simple Explanation

Last digit के (4) non-zero even choices हैं और first digit के (8) non-zero choices बचते हैं। परीक्षा में first और last restrictions को साथ संभालें। / There are (4) non-zero even choices for the last digit and (8) remaining non-zero choices for the first digit. In exams handle first and last restrictions together.

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Digits (1,2,3,4,5,6,7,8) से repetition allowed (6)-digit numbers में exactly (3) even digits हों, तो count क्या है?

Using digits (1,2,3,4,5,6,7,8) with repetition allowed, if exactly (3) even digits occur in (6)-digit numbers, what is the count?

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Correct Answer

A. \(^{6}C_3\cdot4^3\cdot4^3\)

Explanation

Simple Explanation

Even positions चुनें और फिर even तथा odd choices independently multiply करें। परीक्षा में exactly digit type में positions first choose करें। / Choose the even positions and then multiply even and odd choices independently. In exams choose positions first in exactly digit-type problems.

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Length (8) strings (5) symbols से बनती हैं और हर symbol कम से कम एक बार आए। Count का inclusion-exclusion form कौन-सा है?

Length (8) strings are formed from (5) symbols and every symbol appears at least once. Which inclusion-exclusion form gives the count?

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Correct Answer

A. (\sum_{i=0}^{5}(-1)^i{}^{5}C_i(5-i)8)

Explanation

Simple Explanation

हर symbol का आना onto condition है और missing symbols हटते हैं। परीक्षा में at least once को inclusion-exclusion से करें। / Every symbol appearing is an onto condition and missing symbols are removed. In exams solve at least once by inclusion-exclusion.

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Length (9) strings (6) symbols से बनती हैं और exactly (4) distinct symbols use हों। सही count कौन-सी है?

Length (9) strings are formed from (6) symbols and exactly (4) distinct symbols are used. Which count is correct?

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Correct Answer

A. (^{6}C_4\sum_{i=0}^{4}(-1)^i{}^{4}C_i(4-i)9)

Explanation

Simple Explanation

पहले (4) symbols चुनें और फिर उन पर onto strings बनाएं। परीक्षा में exactly distinct symbols में choose set plus onto count करें। / First choose (4) symbols and then form onto strings on them. In exams use choose set plus onto count for exactly distinct symbols.

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Length (r) strings में exactly (s) distinct symbols use हों तो (s!,S(r,s)) किसे count करता है?

In length (r) strings with exactly (s) distinct symbols used, what does (s!,S(r,s)) count?

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Correct Answer

A. (r) positions से selected (s) symbols पर onto assignmentsOnto assignments from (r) positions to selected (s) symbols

Explanation

Simple Explanation

Stirling part positions को non-empty groups में बांटता है और (s!) groups को symbols assign करता है। परीक्षा में exactly used symbols को onto mapping समझें। / The Stirling part partitions positions into non-empty groups and (s!) assigns groups to symbols. In exams treat exactly used symbols as onto mapping.

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((a+b+c+d)^{10}) में \(a^2b^3c^1d^4\) का coefficient क्या है?

What is the coefficient of \(a^2b^3c^1d^4\) in ((a+b+c+d)^{10})?

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Correct Answer

A. \(\frac{10!}{2!3!1!4!}\)

Explanation

Simple Explanation

Exponents का sum (10) है और coefficient multinomial form से मिलता है। परीक्षा में powers को group sizes मानें। / The exponents sum to (10) and the coefficient comes from the multinomial form. In exams treat powers as group sizes.

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(\(1+x+x^2\)^{10}) में \(x^3\) का coefficient किस expression से मिलेगा?

Which expression gives the coefficient of \(x^3\) in (\(1+x+x^2\)^{10})?

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Correct Answer

A. \(^{10}C_3+10\cdot9\)

Explanation

Simple Explanation

Cases हैं: तीन (x) चुनें या एक \(x^2\) और एक (x) चुनें। परीक्षा में same power बनाने वाले all cases जोड़ें। / The cases are: choose three (x)'s or choose one \(x^2\) and one (x). In exams add all cases that form the same power.

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