यदि (5) पुस्तकों में से (2) पुस्तकें चुननी हों तो कुल कितने चयन होंगे?
If (2) books are to be selected from (5) books, how many selections are possible?
#combinations
#class11
#easy
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A (8)
B (10)
C (12)
D (20)
Explanation opens after your attempt
Step 1
Concept
This is a direct use of \(\binom{5}{2}=10\). In exams, order is not counted in selection.
Step 2
Why this answer is correct
The correct answer is B. (10). This is a direct use of \(\binom{5}{2}=10\). In exams, order is not counted in selection.
Step 3
Exam Tip
यह \(\binom{5}{2}=10\) का सरल प्रयोग है। परीक्षा में चयन का क्रम नहीं गिना जाता।
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(6) विद्यार्थियों में से (3) विद्यार्थियों की टीम कितने तरीकों से चुनी जा सकती है?
In how many ways can a team of (3) students be selected from (6) students?
#combinations
#class11
#easy
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A (18)
B (15)
C (20)
D (12)
Explanation opens after your attempt
Step 1
Concept
Order is not important while selecting a team. Hence \(\binom{6}{3}=20\) is correct.
Step 2
Why this answer is correct
The correct answer is C. (20). Order is not important while selecting a team. Hence \(\binom{6}{3}=20\) is correct.
Step 3
Exam Tip
टीम चुनने में क्रम महत्वपूर्ण नहीं होता। इसलिए \(\binom{6}{3}=20\) सही है।
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(7) अलग-अलग फूलों में से (2) फूल चुनने के कितने तरीके हैं?
How many ways are there to choose (2) flowers from (7) different flowers?
#combinations
#class11
#easy
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A (14)
B (21)
C (35)
D (42)
Explanation opens after your attempt
Step 1
Concept
It is \(\binom{7}{2}=\frac{7\cdot6}{2}=21\). Such questions form pairs of objects.
Step 2
Why this answer is correct
The correct answer is B. (21). It is \(\binom{7}{2}=\frac{7\cdot6}{2}=21\). Such questions form pairs of objects.
Step 3
Exam Tip
यह \(\binom{7}{2}=\frac{7\cdot6}{2}=21\) है। ऐसे प्रश्नों में वस्तुओं की जोड़ी बनती है।
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(8) खिलाड़ियों में से (4) खिलाड़ियों का समूह कितने तरीकों से बनाया जा सकता है?
In how many ways can a group of (4) players be formed from (8) players?
#combinations
#class11
#easy
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A (32)
B (56)
C (70)
D (1680)
Explanation opens after your attempt
Step 1
Concept
A group has no order, so \(\binom{8}{4}=70\). In exams, the word group usually indicates combination.
Step 2
Why this answer is correct
The correct answer is C. (70). A group has no order, so \(\binom{8}{4}=70\). In exams, the word group usually indicates combination.
Step 3
Exam Tip
समूह में क्रम नहीं होता इसलिए \(\binom{8}{4}=70\) होगा। परीक्षा में समूह शब्द दिखे तो संयोजन लगाएं।
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(9) प्रश्नों में से (1) प्रश्न चुनने के कितने तरीके हैं?
How many ways are there to select (1) question from (9) questions?
#combinations
#class11
#easy
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A (1)
B (8)
C (9)
D (18)
Explanation opens after your attempt
Step 1
Concept
Selecting one object gives \(\binom{9}{1}=9\). Remember \(\binom{n}{1}=n\).
Step 2
Why this answer is correct
The correct answer is C. (9). Selecting one object gives \(\binom{9}{1}=9\). Remember \(\binom{n}{1}=n\).
Step 3
Exam Tip
एक वस्तु चुनने पर \(\binom{9}{1}=9\) होता है। \(\binom{n}{1}=n\) याद रखें।
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(10) पेन में से (0) पेन चुनने का एक ही तरीका क्यों माना जाता है?
Why is selecting (0) pens from (10) pens considered one way?
#combinations
#class11
#easy
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A क्योंकि \(\binom{10}{0}=1\) / Because \(\binom{10}{0}=1\)
B क्योंकि \(\binom{10}{0}=0\) / Because \(\binom{10}{0}=0\)
C क्योंकि \(\binom{10}{0}=10\) / Because \(\binom{10}{0}=10\)
D क्योंकि \(\binom{10}{0}=100\) / Because \(\binom{10}{0}=100\)
Explanation opens after your attempt
Correct Answer
A. क्योंकि \(\binom{10}{0}=1\) / Because \(\binom{10}{0}=1\)
Step 1
Concept
Choosing nothing is also one definite selection. Therefore \(\binom{n}{0}=1\).
Step 2
Why this answer is correct
The correct answer is A. क्योंकि \(\binom{10}{0}=1\) / Because \(\binom{10}{0}=1\). Choosing nothing is also one definite selection. Therefore \(\binom{n}{0}=1\).
Step 3
Exam Tip
कुछ भी न चुनना भी एक निश्चित चयन है। इसलिए \(\binom{n}{0}=1\) होता है।
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(5) रंगों में से (5) रंग चुनने के कितने तरीके हैं?
How many ways are there to choose all (5) colors from (5) colors?
#combinations
#class11
#easy
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A (0)
B (1)
C (5)
D (25)
Explanation opens after your attempt
Step 1
Concept
There is only one way to choose all objects. Hence \(\binom{5}{5}=1\).
Step 2
Why this answer is correct
The correct answer is B. (1). There is only one way to choose all objects. Hence \(\binom{5}{5}=1\).
Step 3
Exam Tip
सभी वस्तुएं चुनने का केवल एक तरीका होता है। इसलिए \(\binom{5}{5}=1\) है।
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(6) दोस्तों में से (2) दोस्तों की जोड़ी कितने तरीकों से चुनी जा सकती है?
In how many ways can a pair of friends be selected from (6) friends?
#combinations
#class11
#easy
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A (12)
B (15)
C (18)
D (30)
Explanation opens after your attempt
Step 1
Concept
For a pair, \(\binom{6}{2}=15\). In a pair, first and second are not counted separately.
Step 2
Why this answer is correct
The correct answer is B. (15). For a pair, \(\binom{6}{2}=15\). In a pair, first and second are not counted separately.
Step 3
Exam Tip
जोड़ी के लिए \(\binom{6}{2}=15\) होगा। जोड़ी में पहला और दूसरा अलग नहीं गिने जाते।
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(4) लड़कों और (3) लड़कियों में से (1) लड़का और (1) लड़की कितने तरीकों से चुने जा सकते हैं?
From (4) boys and (3) girls, in how many ways can (1) boy and (1) girl be selected?
#combinations
#class11
#easy
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A (7)
B (12)
C (24)
D (1)
Explanation opens after your attempt
Step 1
Concept
The selection is \(\binom{4}{1}\binom{3}{1}=12\). Multiply choices when selecting from different groups.
Step 2
Why this answer is correct
The correct answer is B. (12). The selection is \(\binom{4}{1}\binom{3}{1}=12\). Multiply choices when selecting from different groups.
Step 3
Exam Tip
चयन \(\binom{4}{1}\binom{3}{1}=12\) है। अलग-अलग वर्गों से चयन में गुणा करते हैं।
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(5) लड़कों और (4) लड़कियों में से (2) लड़के चुनने के कितने तरीके हैं?
From (5) boys and (4) girls, how many ways are there to select (2) boys?
#combinations
#class11
#easy
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A (10)
B (20)
C (9)
D (30)
Explanation opens after your attempt
Step 1
Concept
Only boys are being selected, so \(\binom{5}{2}=10\). Do not get confused by extra information.
Step 2
Why this answer is correct
The correct answer is A. (10). Only boys are being selected, so \(\binom{5}{2}=10\). Do not get confused by extra information.
Step 3
Exam Tip
केवल लड़कों में से चयन है इसलिए \(\binom{5}{2}=10\) होगा। अनावश्यक सूचना से भ्रमित न हों।
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(4) सेब और (5) आमों में से (1) सेब और (2) आम कितने तरीकों से चुने जा सकते हैं?
From (4) apples and (5) mangoes, in how many ways can (1) apple and (2) mangoes be selected?
#combinations
#class11
#easy
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A (10)
B (20)
C (40)
D (80)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{4}{1}\binom{5}{2}=4\cdot10=40\). Multiply selections from separate fruit groups.
Step 2
Why this answer is correct
The correct answer is C. (40). The ways are \(\binom{4}{1}\binom{5}{2}=4\cdot10=40\). Multiply selections from separate fruit groups.
Step 3
Exam Tip
तरीके \(\binom{4}{1}\binom{5}{2}=4\cdot10=40\) हैं। अलग फल वर्गों के चयन को गुणा करें।
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(8) बिंदुओं में से (2) बिंदु चुनकर कितनी रेखाएं बनाई जा सकती हैं?
How many lines can be formed by selecting (2) points from (8) points?
#combinations
#class11
#easy
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A (16)
B (28)
C (56)
D (64)
Explanation opens after your attempt
Step 1
Concept
Two points form one line, so \(\binom{8}{2}=28\). The order of points is not counted.
Step 2
Why this answer is correct
The correct answer is B. (28). Two points form one line, so \(\binom{8}{2}=28\). The order of points is not counted.
Step 3
Exam Tip
दो बिंदु एक रेखा बनाते हैं इसलिए \(\binom{8}{2}=28\) है। बिंदुओं का क्रम नहीं गिना जाता।
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(7) बिंदुओं में से कोई (3) एक सीध में नहीं हैं। इनसे कितने त्रिभुज बन सकते हैं?
There are (7) points, no (3) of which are collinear. How many triangles can be formed?
#combinations
#class11
#easy
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A (21)
B (35)
C (42)
D (70)
Explanation opens after your attempt
Step 1
Concept
A triangle needs (3) points, so \(\binom{7}{3}=35\). The non-collinear condition makes triangles possible.
Step 2
Why this answer is correct
The correct answer is B. (35). A triangle needs (3) points, so \(\binom{7}{3}=35\). The non-collinear condition makes triangles possible.
Step 3
Exam Tip
त्रिभुज के लिए (3) बिंदु चाहिए इसलिए \(\binom{7}{3}=35\) है। सीध में न होने की शर्त त्रिभुज को संभव बनाती है।
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(6) बिंदुओं से कितने वृत्त बनाए जा सकते हैं यदि कोई (3) बिंदु एक सीध में नहीं हैं?
How many circles can be formed from (6) points if no (3) points are collinear?
#combinations
#class11
#easy
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A (12)
B (15)
C (20)
D (36)
Explanation opens after your attempt
Step 1
Concept
A circle is determined by (3) points. Hence \(\binom{6}{3}=20\) is correct.
Step 2
Why this answer is correct
The correct answer is C. (20). A circle is determined by (3) points. Hence \(\binom{6}{3}=20\) is correct.
Step 3
Exam Tip
एक वृत्त को (3) बिंदु निर्धारित करते हैं। इसलिए \(\binom{6}{3}=20\) सही है।
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(10) व्यक्तियों में से (2) व्यक्तियों के हाथ मिलाने की कुल संख्या क्या होगी?
What is the total number of handshakes among (10) persons taken (2) at a time?
#combinations
#class11
#easy
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A (20)
B (45)
C (90)
D (100)
Explanation opens after your attempt
Step 1
Concept
Each handshake is a pair of (2) persons. Therefore the total number is \(\binom{10}{2}=45\).
Step 2
Why this answer is correct
The correct answer is B. (45). Each handshake is a pair of (2) persons. Therefore the total number is \(\binom{10}{2}=45\).
Step 3
Exam Tip
हर हाथ मिलाना (2) व्यक्तियों की एक जोड़ी है। इसलिए कुल संख्या \(\binom{10}{2}=45\) है।
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(9) सदस्यों में से (3) सदस्यों की समिति कितने तरीकों से चुनी जा सकती है?
In how many ways can a committee of (3) members be selected from (9) members?
#combinations
#class11
#easy
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A (27)
B (72)
C (84)
D (504)
Explanation opens after your attempt
Step 1
Concept
This is a committee selection question, so \(\binom{9}{3}=84\). If posts are not assigned, order is not counted.
Step 2
Why this answer is correct
The correct answer is C. (84). This is a committee selection question, so \(\binom{9}{3}=84\). If posts are not assigned, order is not counted.
Step 3
Exam Tip
समिति चयन का प्रश्न है इसलिए \(\binom{9}{3}=84\) है। समिति में पद न हों तो क्रम नहीं गिना जाता।
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(12) उम्मीदवारों में से (2) उम्मीदवारों को पुरस्कार के लिए चुनने के कितने तरीके हैं?
How many ways are there to select (2) candidates for an award from (12) candidates?
#combinations
#class11
#easy
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A (24)
B (66)
C (132)
D (144)
Explanation opens after your attempt
Step 1
Concept
The order of the two selected candidates is not important. Thus \(\binom{12}{2}=66\).
Step 2
Why this answer is correct
The correct answer is B. (66). The order of the two selected candidates is not important. Thus \(\binom{12}{2}=66\).
Step 3
Exam Tip
दो समान चयनित उम्मीदवारों का क्रम महत्वपूर्ण नहीं है। अतः \(\binom{12}{2}=66\) होगा।
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(11) अलग-अलग कार्डों में से (3) कार्ड चुनने के कितने तरीके होंगे?
How many ways are there to choose (3) cards from (11) different cards?
#combinations
#class11
#easy
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A (33)
B (99)
C (165)
D (990)
Explanation opens after your attempt
Step 1
Concept
This is a direct use of \(\binom{11}{3}=165\). Order is not considered while choosing cards.
Step 2
Why this answer is correct
The correct answer is C. (165). This is a direct use of \(\binom{11}{3}=165\). Order is not considered while choosing cards.
Step 3
Exam Tip
यह \(\binom{11}{3}=165\) का सीधा प्रयोग है। कार्ड चुनने में क्रम नहीं देखा जाता।
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(6) लाल गेंदों और (4) नीली गेंदों में से (1) लाल और (1) नीली गेंद कितने तरीकों से चुनी जा सकती है?
From (6) red balls and (4) blue balls, in how many ways can (1) red and (1) blue ball be selected?
#combinations
#class11
#easy
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A (10)
B (20)
C (24)
D (30)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{6}{1}\binom{4}{1}=24\). Apply multiplication rule for selections from different colors.
Step 2
Why this answer is correct
The correct answer is C. (24). The ways are \(\binom{6}{1}\binom{4}{1}=24\). Apply multiplication rule for selections from different colors.
Step 3
Exam Tip
तरीके \(\binom{6}{1}\binom{4}{1}=24\) हैं। अलग रंगों के चयन में गुणा नियम लगाएं।
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(5) लाल गेंदों और (3) काली गेंदों में से (2) लाल गेंदें चुनने के कितने तरीके हैं?
From (5) red balls and (3) black balls, how many ways are there to choose (2) red balls?
#combinations
#class11
#easy
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A (8)
B (10)
C (15)
D (30)
Explanation opens after your attempt
Step 1
Concept
Only red balls are being selected. Hence \(\binom{5}{2}=10\) is correct.
Step 2
Why this answer is correct
The correct answer is B. (10). Only red balls are being selected. Hence \(\binom{5}{2}=10\) is correct.
Step 3
Exam Tip
केवल लाल गेंदों में से चयन है। इसलिए \(\binom{5}{2}=10\) सही है।
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(7) कुर्सियों में से (3) कुर्सियां चुनने के कितने तरीके हैं?
How many ways are there to select (3) chairs from (7) chairs?
#combinations
#class11
#easy
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A (21)
B (35)
C (42)
D (210)
Explanation opens after your attempt
Step 1
Concept
No arrangement is required, so \(\binom{7}{3}=35\). Choosing chairs is a combination.
Step 2
Why this answer is correct
The correct answer is B. (35). No arrangement is required, so \(\binom{7}{3}=35\). Choosing chairs is a combination.
Step 3
Exam Tip
चयन में व्यवस्था नहीं करनी है इसलिए \(\binom{7}{3}=35\) है। कुर्सियां चुनना संयोजन है।
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(13) विद्यार्थियों में से कक्षा प्रतिनिधि समिति के लिए (2) विद्यार्थी कितने तरीकों से चुने जा सकते हैं?
In how many ways can (2) students be selected from (13) students for a class representative committee?
#combinations
#class11
#easy
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A (26)
B (78)
C (156)
D (169)
Explanation opens after your attempt
Step 1
Concept
It is only selection for a committee, so \(\binom{13}{2}=78\). If posts are not different, order is not counted.
Step 2
Why this answer is correct
The correct answer is B. (78). It is only selection for a committee, so \(\binom{13}{2}=78\). If posts are not different, order is not counted.
Step 3
Exam Tip
समिति के लिए केवल चयन है इसलिए \(\binom{13}{2}=78\) होगा। पद अलग न हों तो क्रम नहीं गिना जाता।
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(6) विषयों में से (4) विषय चुनने के कितने तरीके हैं?
How many ways are there to choose (4) subjects from (6) subjects?
#combinations
#class11
#easy
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A (10)
B (12)
C (15)
D (24)
Explanation opens after your attempt
Step 1
Concept
\(\binom{6}{4}=\binom{6}{2}=15\). Remember \(\binom{n}{r}=\binom{n}{n-r}\).
Step 2
Why this answer is correct
The correct answer is C. (15). \(\binom{6}{4}=\binom{6}{2}=15\). Remember \(\binom{n}{r}=\binom{n}{n-r}\).
Step 3
Exam Tip
\(\binom{6}{4}=\binom{6}{2}=15\) होता है। \(\binom{n}{r}=\binom{n}{n-r}\) याद रखें।
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\(\binom{8}{2}\) का मान क्या है?
What is the value of \(\binom{8}{2}\)?
#combinations
#class11
#easy
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A (16)
B (24)
C (28)
D (56)
Explanation opens after your attempt
Step 1
Concept
\(\binom{8}{2}=\frac{8\cdot7}{2}=28\). The formula for choosing two objects is quick.
Step 2
Why this answer is correct
The correct answer is C. (28). \(\binom{8}{2}=\frac{8\cdot7}{2}=28\). The formula for choosing two objects is quick.
Step 3
Exam Tip
\(\binom{8}{2}=\frac{8\cdot7}{2}=28\) है। दो वस्तुओं के चयन का सूत्र तेज होता है।
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\(\binom{10}{3}\) का मान निकालिए।
Find the value of \(\binom{10}{3}\).
#combinations
#class11
#easy
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A (30)
B (90)
C (120)
D (720)
Explanation opens after your attempt
Step 1
Concept
\(\binom{10}{3}=\frac{10\cdot9\cdot8}{3\cdot2\cdot1}=120\). For small (r), multiply directly.
Step 2
Why this answer is correct
The correct answer is C. (120). \(\binom{10}{3}=\frac{10\cdot9\cdot8}{3\cdot2\cdot1}=120\). For small (r), multiply directly.
Step 3
Exam Tip
\(\binom{10}{3}=\frac{10\cdot9\cdot8}{3\cdot2\cdot1}=120\) है। छोटे (r) के लिए सीधे गुणा करें।
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\(\binom{9}{0}\) का मान क्या होगा?
What will be the value of \(\binom{9}{0}\)?
#combinations
#class11
#easy
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A (0)
B (1)
C (9)
D (90)
Explanation opens after your attempt
Step 1
Concept
For any (n), \(\binom{n}{0}=1\). This is a basic identity of combinations.
Step 2
Why this answer is correct
The correct answer is B. (1). For any (n), \(\binom{n}{0}=1\). This is a basic identity of combinations.
Step 3
Exam Tip
किसी भी (n) के लिए \(\binom{n}{0}=1\) होता है। यह संयोजन की मूल पहचान है।
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\(\binom{7}{7}\) का मान क्या है?
What is the value of \(\binom{7}{7}\)?
#combinations
#class11
#easy
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A (0)
B (1)
C (7)
D (49)
Explanation opens after your attempt
Step 1
Concept
There is only one way to choose all (7) objects. Therefore \(\binom{7}{7}=1\).
Step 2
Why this answer is correct
The correct answer is B. (1). There is only one way to choose all (7) objects. Therefore \(\binom{7}{7}=1\).
Step 3
Exam Tip
सभी (7) वस्तुओं को चुनने का केवल एक तरीका है। अतः \(\binom{7}{7}=1\) है।
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\(\binom{12}{1}\) का मान क्या है?
What is the value of \(\binom{12}{1}\)?
#combinations
#class11
#easy
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A (1)
B (11)
C (12)
D (24)
Explanation opens after your attempt
Step 1
Concept
\(\binom{n}{1}=n\). Hence \(\binom{12}{1}=12\) is correct.
Step 2
Why this answer is correct
The correct answer is C. (12). \(\binom{n}{1}=n\). Hence \(\binom{12}{1}=12\) is correct.
Step 3
Exam Tip
\(\binom{n}{1}=n\) होता है। इसलिए \(\binom{12}{1}=12\) सही है।
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\(\binom{9}{2}\) का मान ज्ञात कीजिए।
Find the value of \(\binom{9}{2}\).
#combinations
#class11
#easy
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A (18)
B (36)
C (72)
D (81)
Explanation opens after your attempt
Step 1
Concept
\(\binom{9}{2}=\frac{9\cdot8}{2}=36\). Do not forget to divide by (2!) when choosing two objects.
Step 2
Why this answer is correct
The correct answer is B. (36). \(\binom{9}{2}=\frac{9\cdot8}{2}=36\). Do not forget to divide by (2!) when choosing two objects.
Step 3
Exam Tip
\(\binom{9}{2}=\frac{9\cdot8}{2}=36\) है। दो वस्तुओं के चयन में (2!) से भाग देना न भूलें।
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\(\binom{11}{1}+\binom{11}{0}\) का मान क्या है?
What is the value of \(\binom{11}{1}+\binom{11}{0}\)?
#combinations
#class11
#easy
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A (10)
B (11)
C (12)
D (22)
Explanation opens after your attempt
Step 1
Concept
The values are \(\binom{11}{1}=11\) and \(\binom{11}{0}=1\). The total is (12).
Step 2
Why this answer is correct
The correct answer is C. (12). The values are \(\binom{11}{1}=11\) and \(\binom{11}{0}=1\). The total is (12).
Step 3
Exam Tip
मान \(\binom{11}{1}=11\) और \(\binom{11}{0}=1\) हैं। कुल (12) होगा।
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यदि \(\binom{n}{1}=14\) है तो (n) का मान क्या है?
If \(\binom{n}{1}=14\), what is the value of (n)?
#combinations
#class11
#easy
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A (7)
B (13)
C (14)
D (28)
Explanation opens after your attempt
Step 1
Concept
\(\binom{n}{1}=n\). Therefore (n=14).
Step 2
Why this answer is correct
The correct answer is C. (14). \(\binom{n}{1}=n\). Therefore (n=14).
Step 3
Exam Tip
\(\binom{n}{1}=n\) होता है। इसलिए (n=14) मिलेगा।
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यदि \(\binom{n}{0}=1\) है तो यह किस (n) के लिए सत्य है?
If \(\binom{n}{0}=1\), for which (n) is it true?
#combinations
#class11
#easy
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A केवल (n=1) / Only (n=1)
B केवल (n=0) / Only (n=0)
C हर \(n\geq0\) / Every \(n\geq0\)
D कोई (n) नहीं / No (n)
Explanation opens after your attempt
Correct Answer
C. हर \(n\geq0\) / Every \(n\geq0\)
Step 1
Concept
There is one way to choose zero objects for every \(n\geq0\). This is a basic rule.
Step 2
Why this answer is correct
The correct answer is C. हर \(n\geq0\) / Every \(n\geq0\). There is one way to choose zero objects for every \(n\geq0\). This is a basic rule.
Step 3
Exam Tip
शून्य वस्तु चुनने का एक तरीका हर \(n\geq0\) के लिए होता है। यह मूल नियम है।
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\(\binom{6}{2}\) और \(\binom{6}{4}\) में क्या संबंध है?
What is the relation between \(\binom{6}{2}\) and \(\binom{6}{4}\)?
#combinations
#class11
#easy
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A दोनों बराबर हैं / They are equal
B पहला दुगुना है / The first is double
C दूसरा शून्य है / The second is zero
D दोनों का योग (6) है / Their sum is (6)
Explanation opens after your attempt
Correct Answer
A. दोनों बराबर हैं / They are equal
Step 1
Concept
Because \(\binom{n}{r}=\binom{n}{n-r}\). Here (r=2) and (n-r=4).
Step 2
Why this answer is correct
The correct answer is A. दोनों बराबर हैं / They are equal. Because \(\binom{n}{r}=\binom{n}{n-r}\). Here (r=2) and (n-r=4).
Step 3
Exam Tip
क्योंकि \(\binom{n}{r}=\binom{n}{n-r}\) होता है। यहां (r=2) और (n-r=4) हैं।
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\(\binom{5}{2}+\binom{5}{3}\) का मान क्या है?
What is the value of \(\binom{5}{2}+\binom{5}{3}\)?
#combinations
#class11
#easy
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A (10)
B (15)
C (20)
D (25)
Explanation opens after your attempt
Step 1
Concept
\(\binom{5}{2}=10\) and \(\binom{5}{3}=10\). Therefore the total is (20).
Step 2
Why this answer is correct
The correct answer is C. (20). \(\binom{5}{2}=10\) and \(\binom{5}{3}=10\). Therefore the total is (20).
Step 3
Exam Tip
\(\binom{5}{2}=10\) और \(\binom{5}{3}=10\) हैं। इसलिए कुल (20) है।
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(3) गणित और (4) विज्ञान पुस्तकों में से (1) गणित और (1) विज्ञान पुस्तक कितने तरीकों से चुनी जा सकती है?
From (3) mathematics books and (4) science books, in how many ways can (1) mathematics book and (1) science book be chosen?
#combinations
#class11
#easy
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A (7)
B (12)
C (24)
D (1)
Explanation opens after your attempt
Step 1
Concept
The selection is \(\binom{3}{1}\binom{4}{1}=12\). Choices from different types are multiplied.
Step 2
Why this answer is correct
The correct answer is B. (12). The selection is \(\binom{3}{1}\binom{4}{1}=12\). Choices from different types are multiplied.
Step 3
Exam Tip
चयन \(\binom{3}{1}\binom{4}{1}=12\) है। अलग प्रकारों में चयन की संख्याएं गुणा होती हैं।
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(5) पेन और (6) पेंसिलों में से (2) पेन और (1) पेंसिल कितने तरीकों से चुनी जा सकती हैं?
From (5) pens and (6) pencils, in how many ways can (2) pens and (1) pencil be selected?
#combinations
#class11
#easy
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A (30)
B (45)
C (60)
D (120)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{5}{2}\binom{6}{1}=10\cdot6=60\). Select from each group separately.
Step 2
Why this answer is correct
The correct answer is C. (60). The ways are \(\binom{5}{2}\binom{6}{1}=10\cdot6=60\). Select from each group separately.
Step 3
Exam Tip
तरीके \(\binom{5}{2}\binom{6}{1}=10\cdot6=60\) हैं। प्रत्येक समूह से चयन अलग-अलग करें।
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(4) सफेद और (5) काली गेंदों में से (2) काली गेंदें कितने तरीकों से चुनी जा सकती हैं?
From (4) white and (5) black balls, in how many ways can (2) black balls be selected?
#combinations
#class11
#easy
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A (10)
B (20)
C (25)
D (40)
Explanation opens after your attempt
Step 1
Concept
Only black balls are selected, so \(\binom{5}{2}=10\). The remaining information is only context.
Step 2
Why this answer is correct
The correct answer is A. (10). Only black balls are selected, so \(\binom{5}{2}=10\). The remaining information is only context.
Step 3
Exam Tip
केवल काली गेंदों में से चयन है इसलिए \(\binom{5}{2}=10\) होगा। बाकी सूचना केवल संदर्भ है।
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(8) छात्रों में से (2) छात्रों को पुस्तकालय कार्य के लिए चुनना है। कितने चयन संभव हैं?
(2) students are to be selected from (8) students for library work. How many selections are possible?
#combinations
#class11
#easy
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A (16)
B (28)
C (56)
D (64)
Explanation opens after your attempt
Step 1
Concept
This is \(\binom{8}{2}=28\). If the work is same, order is not counted in selection.
Step 2
Why this answer is correct
The correct answer is B. (28). This is \(\binom{8}{2}=28\). If the work is same, order is not counted in selection.
Step 3
Exam Tip
यह \(\binom{8}{2}=28\) है। कार्य समान हो तो चयन में क्रम नहीं गिना जाता।
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(6) विद्यार्थियों में से (5) विद्यार्थियों को प्रतियोगिता के लिए चुनने के कितने तरीके हैं?
How many ways are there to select (5) students from (6) students for a competition?
#combinations
#class11
#easy
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A (5)
B (6)
C (15)
D (30)
Explanation opens after your attempt
Step 1
Concept
\(\binom{6}{5}=\binom{6}{1}=6\). Sometimes counting the person not selected is easier.
Step 2
Why this answer is correct
The correct answer is B. (6). \(\binom{6}{5}=\binom{6}{1}=6\). Sometimes counting the person not selected is easier.
Step 3
Exam Tip
\(\binom{6}{5}=\binom{6}{1}=6\) है। कभी-कभी न चुने गए व्यक्ति को गिनना आसान होता है।
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(10) अलग-अलग प्रश्नों में से (3) प्रश्नों का सेट कितने तरीकों से बनाया जा सकता है?
In how many ways can a set of (3) questions be made from (10) different questions?
#combinations
#class11
#easy
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A (30)
B (90)
C (120)
D (720)
Explanation opens after your attempt
Step 1
Concept
Making a set of questions is selection. Hence \(\binom{10}{3}=120\) is correct.
Step 2
Why this answer is correct
The correct answer is C. (120). Making a set of questions is selection. Hence \(\binom{10}{3}=120\) is correct.
Step 3
Exam Tip
प्रश्नों का सेट बनाना चयन है। इसलिए \(\binom{10}{3}=120\) सही है।
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(5) व्यंजनों में से (2) व्यंजन चुनकर थाली बनानी है। कितनी थालियां बन सकती हैं?
A plate is to be made by choosing (2) dishes from (5) dishes. How many plates can be made?
#combinations
#class11
#easy
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A (10)
B (20)
C (25)
D (5)
Explanation opens after your attempt
Step 1
Concept
The ways to choose two dishes are \(\binom{5}{2}=10\). The order on the plate is not important.
Step 2
Why this answer is correct
The correct answer is A. (10). The ways to choose two dishes are \(\binom{5}{2}=10\). The order on the plate is not important.
Step 3
Exam Tip
दो व्यंजन चुनने के तरीके \(\binom{5}{2}=10\) हैं। थाली में क्रम का महत्व नहीं है।
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(7) अध्यापकों में से (2) अध्यापकों की निगरानी टीम कितने तरीकों से चुनी जा सकती है?
In how many ways can an invigilation team of (2) teachers be selected from (7) teachers?
#combinations
#class11
#easy
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A (14)
B (21)
C (42)
D (49)
Explanation opens after your attempt
Step 1
Concept
For the team, there are \(\binom{7}{2}=21\) selections. A team has no order.
Step 2
Why this answer is correct
The correct answer is B. (21). For the team, there are \(\binom{7}{2}=21\) selections. A team has no order.
Step 3
Exam Tip
टीम के लिए \(\binom{7}{2}=21\) चयन होंगे। टीम में क्रम नहीं होता।
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(9) गीतों में से (4) गीत चुनने के कितने तरीके हैं?
How many ways are there to choose (4) songs from (9) songs?
#combinations
#class11
#easy
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A (36)
B (72)
C (126)
D (3024)
Explanation opens after your attempt
Step 1
Concept
Choosing songs is not arrangement, so \(\binom{9}{4}=126\). If order were needed, it would be a different question.
Step 2
Why this answer is correct
The correct answer is C. (126). Choosing songs is not arrangement, so \(\binom{9}{4}=126\). If order were needed, it would be a different question.
Step 3
Exam Tip
गीत चुनना व्यवस्था नहीं है इसलिए \(\binom{9}{4}=126\) है। क्रम चाहिए होता तो अलग प्रश्न होता।
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(8) अलग-अलग चित्रों में से (3) चित्र प्रदर्शनी के लिए चुनने के कितने तरीके हैं?
How many ways are there to choose (3) pictures from (8) different pictures for an exhibition?
#combinations
#class11
#easy
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A (24)
B (56)
C (112)
D (336)
Explanation opens after your attempt
Step 1
Concept
The selection is \(\binom{8}{3}=56\). Choosing for an exhibition is a combination.
Step 2
Why this answer is correct
The correct answer is B. (56). The selection is \(\binom{8}{3}=56\). Choosing for an exhibition is a combination.
Step 3
Exam Tip
चयन \(\binom{8}{3}=56\) होगा। प्रदर्शनी के लिए चुनना संयोजन है।
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(6) अंकों में से (2) अंक चुनने के कितने तरीके हैं यदि क्रम नहीं देखना है?
How many ways are there to choose (2) digits from (6) digits if order is not considered?
#combinations
#class11
#easy
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A (12)
B (15)
C (30)
D (36)
Explanation opens after your attempt
Step 1
Concept
Order is not considered, so \(\binom{6}{2}=15\). This is the main difference between combination and permutation.
Step 2
Why this answer is correct
The correct answer is B. (15). Order is not considered, so \(\binom{6}{2}=15\). This is the main difference between combination and permutation.
Step 3
Exam Tip
क्रम नहीं देखना है इसलिए \(\binom{6}{2}=15\) है। यही संयोजन और क्रमचय का मुख्य अंतर है।
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(10) बिंदुओं से कितने रेखाखंड बनाए जा सकते हैं?
How many line segments can be formed from (10) points?
#combinations
#class11
#easy
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A (20)
B (45)
C (90)
D (100)
Explanation opens after your attempt
Step 1
Concept
A line segment needs (2) points. Therefore \(\binom{10}{2}=45\).
Step 2
Why this answer is correct
The correct answer is B. (45). A line segment needs (2) points. Therefore \(\binom{10}{2}=45\).
Step 3
Exam Tip
एक रेखाखंड के लिए (2) बिंदु चाहिए। इसलिए \(\binom{10}{2}=45\) है।
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(8) अक्षरों में से (3) अक्षरों का चयन कितने तरीकों से हो सकता है यदि क्रम महत्वपूर्ण नहीं है?
In how many ways can (3) letters be selected from (8) letters if order is not important?
#combinations
#class11
#easy
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A (24)
B (56)
C (336)
D (512)
Explanation opens after your attempt
Step 1
Concept
Order is not important, so \(\binom{8}{3}=56\). Use combinations in selection questions.
Step 2
Why this answer is correct
The correct answer is B. (56). Order is not important, so \(\binom{8}{3}=56\). Use combinations in selection questions.
Step 3
Exam Tip
क्रम महत्वपूर्ण नहीं है इसलिए \(\binom{8}{3}=56\) है। चयन वाले प्रश्नों में संयोजन लगाएं।
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(5) लड़कों और (5) लड़कियों में से (1) लड़का और (1) लड़की चुनने के कितने तरीके हैं?
From (5) boys and (5) girls, in how many ways can (1) boy and (1) girl be selected?
#combinations
#class11
#easy
50 50-50 2 wrong hide
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A (10)
B (20)
C (25)
D (50)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{5}{1}\binom{5}{1}=25\). The two selections are independent, so multiply.
Step 2
Why this answer is correct
The correct answer is C. (25). The ways are \(\binom{5}{1}\binom{5}{1}=25\). The two selections are independent, so multiply.
Step 3
Exam Tip
तरीके \(\binom{5}{1}\binom{5}{1}=25\) हैं। दोनों चयन स्वतंत्र हैं इसलिए गुणा करें।
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(12) खिलौनों में से (4) खिलौने चुनने के कितने तरीके हैं?
How many ways are there to choose (4) toys from (12) toys?
#combinations
#class11
#easy
50 50-50 2 wrong hide
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A (48)
B (120)
C (495)
D (11880)
Explanation opens after your attempt
Step 1
Concept
This is \(\binom{12}{4}=495\). For larger values, simplify the formula step by step.
Step 2
Why this answer is correct
The correct answer is C. (495). This is \(\binom{12}{4}=495\). For larger values, simplify the formula step by step.
Step 3
Exam Tip
यह \(\binom{12}{4}=495\) है। बड़े मानों में सूत्र को क्रम से सरल करें।
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(7) रंगीन पेंसिलों में से कम से कम (1) पेंसिल चुनने के कितने तरीके हैं?
In how many ways can at least (1) pencil be selected from (7) colored pencils?
#combinations
#class11
#easy
50 50-50 2 wrong hide
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+10 Time+ 10 sec extra
? Hint Small clue
A (7)
B (49)
C (127)
D (128)
Explanation opens after your attempt
Step 1
Concept
Total subsets are \(2^7\), and the empty selection is removed. Hence the ways are \(2^7-1=127\).
Step 2
Why this answer is correct
The correct answer is C. (127). Total subsets are \(2^7\), and the empty selection is removed. Hence the ways are \(2^7-1=127\).
Step 3
Exam Tip
कुल उपसमुच्चय \(2^7\) हैं और खाली चयन हटेगा। इसलिए \(2^7-1=127\) तरीके हैं।
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