(15) विद्यार्थियों में से (2) विद्यार्थियों की जोड़ी चुनने के कितने तरीके हैं?
How many ways are there to select a pair of (2) students from (15) students?
#combinations
#class11
#easy
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A (105)
B (30)
C (210)
D (225)
Explanation opens after your attempt
Step 1
Concept
The number of ways is \(\binom{15}{2}=105\). Order is not counted in a pair.
Step 2
Why this answer is correct
The correct answer is A. (105). The number of ways is \(\binom{15}{2}=105\). Order is not counted in a pair.
Step 3
Exam Tip
जोड़ी चुनने के तरीके \(\binom{15}{2}=105\) हैं। जोड़ी में क्रम नहीं गिना जाता।
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(8) विद्यार्थियों में से (6) विद्यार्थियों की टीम कितने तरीकों से चुनी जा सकती है?
In how many ways can a team of (6) students be selected from (8) students?
#combinations
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#easy
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A (14)
B (28)
C (48)
D (56)
Explanation opens after your attempt
Step 1
Concept
\(\binom{8}{6}=\binom{8}{2}=28\). When many are selected, counting those not selected is easier.
Step 2
Why this answer is correct
The correct answer is B. (28). \(\binom{8}{6}=\binom{8}{2}=28\). When many are selected, counting those not selected is easier.
Step 3
Exam Tip
\(\binom{8}{6}=\binom{8}{2}=28\) होता है। अधिक चुनना हो तो न चुने गए लोगों को गिनना आसान होता है।
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(9) पुस्तकों में से (3) पुस्तकों का सेट कितने तरीकों से बनाया जा सकता है?
In how many ways can a set of (3) books be made from (9) books?
#combinations
#class11
#easy
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A (27)
B (72)
C (84)
D (504)
Explanation opens after your attempt
Step 1
Concept
Order is not important while making a set. Hence \(\binom{9}{3}=84\) is correct.
Step 2
Why this answer is correct
The correct answer is C. (84). Order is not important while making a set. Hence \(\binom{9}{3}=84\) is correct.
Step 3
Exam Tip
सेट बनाने में क्रम का महत्व नहीं होता। इसलिए \(\binom{9}{3}=84\) सही है।
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(11) खिलाड़ियों में से (4) खिलाड़ियों का अभ्यास समूह कितने तरीकों से चुना जा सकता है?
In how many ways can a practice group of (4) players be selected from (11) players?
#combinations
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#easy
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A (44)
B (165)
C (330)
D (7920)
Explanation opens after your attempt
Step 1
Concept
This is selection, so the value is \(\binom{11}{4}=330\). The correct option is (330).
Step 2
Why this answer is correct
The correct answer is B. (165). This is selection, so the value is \(\binom{11}{4}=330\). The correct option is (330).
Step 3
Exam Tip
यह चयन है इसलिए \(\binom{11}{4}=330\) नहीं बल्कि सही मान \(\binom{11}{4}=330\) है। विकल्पों में सही उत्तर (330) है।
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(12) पेन में से (3) पेन चुनने के कितने तरीके हैं?
How many ways are there to choose (3) pens from (12) pens?
#combinations
#class11
#easy
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A (36)
B (132)
C (220)
D (440)
Explanation opens after your attempt
Step 1
Concept
The number of ways is \(\binom{12}{3}=220\). The order of pens is not considered in selection.
Step 2
Why this answer is correct
The correct answer is C. (220). The number of ways is \(\binom{12}{3}=220\). The order of pens is not considered in selection.
Step 3
Exam Tip
तीन पेन चुनने के तरीके \(\binom{12}{3}=220\) हैं। चयन में पेन का क्रम नहीं देखा जाता।
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(10) प्रश्नों में से (7) प्रश्न चुनने के कितने तरीके हैं?
How many ways are there to select (7) questions from (10) questions?
#combinations
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#easy
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A (120)
B (210)
C (5040)
D (70)
Explanation opens after your attempt
Step 1
Concept
\(\binom{10}{7}=\binom{10}{3}=120\). Complementary selection makes calculation faster.
Step 2
Why this answer is correct
The correct answer is A. (120). \(\binom{10}{7}=\binom{10}{3}=120\). Complementary selection makes calculation faster.
Step 3
Exam Tip
\(\binom{10}{7}=\binom{10}{3}=120\) है। पूरक चयन से गणना जल्दी होती है।
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(5) लड़कों और (4) लड़कियों में से (2) लड़के और (2) लड़कियां कितने तरीकों से चुनी जा सकती हैं?
From (5) boys and (4) girls, in how many ways can (2) boys and (2) girls be selected?
#combinations
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#easy
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A (20)
B (40)
C (60)
D (120)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{5}{2}\binom{4}{2}=10\cdot6=60\). Multiply selections from different groups.
Step 2
Why this answer is correct
The correct answer is C. (60). The ways are \(\binom{5}{2}\binom{4}{2}=10\cdot6=60\). Multiply selections from different groups.
Step 3
Exam Tip
तरीके \(\binom{5}{2}\binom{4}{2}=10\cdot6=60\) हैं। अलग वर्गों से चयन में गुणा करते हैं।
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(6) लाल और (5) नीली गेंदों में से (2) लाल और (1) नीली गेंद कितने तरीकों से चुनी जा सकती है?
From (6) red and (5) blue balls, in how many ways can (2) red and (1) blue ball be selected?
#combinations
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#easy
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A (30)
B (45)
C (60)
D (75)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{6}{2}\binom{5}{1}=15\cdot5=75\). Select separately when colors are different.
Step 2
Why this answer is correct
The correct answer is D. (75). The ways are \(\binom{6}{2}\binom{5}{1}=15\cdot5=75\). Select separately when colors are different.
Step 3
Exam Tip
तरीके \(\binom{6}{2}\binom{5}{1}=15\cdot5=75\) हैं। रंग अलग हों तो चयन अलग-अलग करें।
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(7) पुरुषों और (3) महिलाओं में से (1) पुरुष और (2) महिलाएं कितने तरीकों से चुनी जा सकती हैं?
From (7) men and (3) women, in how many ways can (1) man and (2) women be selected?
#combinations
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#easy
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A (10)
B (21)
C (42)
D (63)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{7}{1}\binom{3}{2}=7\cdot3=21\). Write each condition as a separate combination.
Step 2
Why this answer is correct
The correct answer is B. (21). The ways are \(\binom{7}{1}\binom{3}{2}=7\cdot3=21\). Write each condition as a separate combination.
Step 3
Exam Tip
तरीके \(\binom{7}{1}\binom{3}{2}=7\cdot3=21\) हैं। प्रत्येक शर्त को अलग संयोजन से लिखें।
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(4) गणित और (5) भौतिकी पुस्तकों में से (1) गणित और (2) भौतिकी पुस्तकें कितने तरीकों से चुनी जा सकती हैं?
From (4) mathematics and (5) physics books, in how many ways can (1) mathematics book and (2) physics books be chosen?
#combinations
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#easy
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A (20)
B (30)
C (40)
D (60)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{4}{1}\binom{5}{2}=4\cdot10=40\). The multiplication rule applies to selections from different subjects.
Step 2
Why this answer is correct
The correct answer is C. (40). The ways are \(\binom{4}{1}\binom{5}{2}=4\cdot10=40\). The multiplication rule applies to selections from different subjects.
Step 3
Exam Tip
तरीके \(\binom{4}{1}\binom{5}{2}=4\cdot10=40\) हैं। अलग विषयों के चयन में गुणा नियम लगता है।
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(13) बिंदुओं में से (2) बिंदु चुनकर कितने रेखाखंड बनाए जा सकते हैं?
How many line segments can be formed by choosing (2) points from (13) points?
#combinations
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#easy
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A (26)
B (39)
C (78)
D (156)
Explanation opens after your attempt
Step 1
Concept
A line segment needs (2) points. Therefore \(\binom{13}{2}=78\).
Step 2
Why this answer is correct
The correct answer is C. (78). A line segment needs (2) points. Therefore \(\binom{13}{2}=78\).
Step 3
Exam Tip
एक रेखाखंड के लिए (2) बिंदु चाहिए। अतः \(\binom{13}{2}=78\) होगा।
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(8) बिंदुओं में से कोई (3) एक सीध में नहीं हैं। इनसे कितने त्रिभुज बन सकते हैं?
There are (8) points, no (3) of which are collinear. How many triangles can be formed?
#combinations
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#easy
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A (24)
B (56)
C (112)
D (336)
Explanation opens after your attempt
Step 1
Concept
A triangle is formed by choosing (3) points. Hence \(\binom{8}{3}=56\).
Step 2
Why this answer is correct
The correct answer is B. (56). A triangle is formed by choosing (3) points. Hence \(\binom{8}{3}=56\).
Step 3
Exam Tip
त्रिभुज के लिए (3) बिंदु चुने जाते हैं। इसलिए \(\binom{8}{3}=56\) है।
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(9) बिंदुओं में से कोई (3) एक सीध में नहीं हैं। कितने वृत्त बन सकते हैं?
There are (9) points, no (3) of which are collinear. How many circles can be formed?
#combinations
#class11
#easy
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A (27)
B (72)
C (84)
D (126)
Explanation opens after your attempt
Step 1
Concept
A circle is determined by (3) points. Hence \(\binom{9}{3}=84\) is correct.
Step 2
Why this answer is correct
The correct answer is C. (84). A circle is determined by (3) points. Hence \(\binom{9}{3}=84\) is correct.
Step 3
Exam Tip
एक वृत्त (3) बिंदुओं से निर्धारित होता है। इसलिए \(\binom{9}{3}=84\) सही है।
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(14) व्यक्तियों में से (2) व्यक्तियों के हाथ मिलाने की कुल संख्या क्या होगी?
What is the total number of handshakes among (14) persons?
#combinations
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#easy
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A (28)
B (91)
C (182)
D (196)
Explanation opens after your attempt
Step 1
Concept
Each handshake is a pair of (2) persons. The total is \(\binom{14}{2}=91\).
Step 2
Why this answer is correct
The correct answer is B. (91). Each handshake is a pair of (2) persons. The total is \(\binom{14}{2}=91\).
Step 3
Exam Tip
प्रत्येक हाथ मिलाना (2) व्यक्तियों की जोड़ी है। कुल \(\binom{14}{2}=91\) होंगे।
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(16) टीमों के टूर्नामेंट में प्रत्येक टीम हर दूसरी टीम से एक बार खेले तो कुल मैच कितने होंगे?
In a tournament of (16) teams, if each team plays every other team once, how many matches will be played?
#combinations
#class11
#easy
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A (120)
B (160)
C (240)
D (256)
Explanation opens after your attempt
Step 1
Concept
Each match is a pair of (2) teams. Hence there will be \(\binom{16}{2}=120\) matches.
Step 2
Why this answer is correct
The correct answer is A. (120). Each match is a pair of (2) teams. Hence there will be \(\binom{16}{2}=120\) matches.
Step 3
Exam Tip
प्रत्येक मैच (2) टीमों की जोड़ी है। इसलिए \(\binom{16}{2}=120\) मैच होंगे।
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(6) सदस्यों में से अध्यक्ष या सचिव बनाए बिना (3) सदस्यों की समिति कितने तरीकों से बनेगी?
In how many ways can a committee of (3) members be formed from (6) members without assigning president or secretary?
#combinations
#class11
#easy
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A (18)
B (20)
C (120)
D (216)
Explanation opens after your attempt
Step 1
Concept
No posts are assigned, so it is a combination. \(\binom{6}{3}=20\).
Step 2
Why this answer is correct
The correct answer is B. (20). No posts are assigned, so it is a combination. \(\binom{6}{3}=20\).
Step 3
Exam Tip
पद नहीं दिए गए हैं इसलिए यह संयोजन है। \(\binom{6}{3}=20\) होगा।
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(10) सदस्यों में से (4) सदस्यों की समिति बनानी है जिसमें एक निश्चित सदस्य अवश्य हो। कितने तरीके हैं?
A committee of (4) members is to be formed from (10) members with one fixed member included. How many ways are there?
#combinations
#class11
#easy
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A (84)
B (126)
C (210)
D (5040)
Explanation opens after your attempt
Step 1
Concept
The fixed member is already selected, so choose the remaining (3) from (9). The ways are \(\binom{9}{3}=84\).
Step 2
Why this answer is correct
The correct answer is A. (84). The fixed member is already selected, so choose the remaining (3) from (9). The ways are \(\binom{9}{3}=84\).
Step 3
Exam Tip
निश्चित सदस्य पहले से चुना है इसलिए बाकी (3) सदस्य (9) में से चुनेंगे। तरीके \(\binom{9}{3}=84\) हैं।
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(9) सदस्यों में से (3) सदस्यों की समिति बनानी है जिसमें एक निश्चित सदस्य शामिल न हो। कितने तरीके हैं?
A committee of (3) members is to be formed from (9) members excluding one fixed member. How many ways are there?
#combinations
#class11
#easy
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A (56)
B (84)
C (168)
D (24)
Explanation opens after your attempt
Step 1
Concept
After excluding one member, (8) members remain. Therefore there are \(\binom{8}{3}=56\) ways.
Step 2
Why this answer is correct
The correct answer is A. (56). After excluding one member, (8) members remain. Therefore there are \(\binom{8}{3}=56\) ways.
Step 3
Exam Tip
एक सदस्य हटाने पर (8) सदस्य बचते हैं। इसलिए \(\binom{8}{3}=56\) तरीके हैं।
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(7) विद्यार्थियों में से (3) विद्यार्थियों को चुनना है और दो विशेष विद्यार्थी दोनों साथ में चुने जाने चाहिए। कितने तरीके हैं?
From (7) students, (3) students are to be selected and two special students must be selected together. How many ways are there?
#combinations
#class11
#easy
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A (3)
B (5)
C (10)
D (21)
Explanation opens after your attempt
Step 1
Concept
The two special students are already selected. The third student is chosen from the remaining (5) in \(\binom{5}{1}=5\) ways.
Step 2
Why this answer is correct
The correct answer is B. (5). The two special students are already selected. The third student is chosen from the remaining (5) in \(\binom{5}{1}=5\) ways.
Step 3
Exam Tip
दो विशेष विद्यार्थी पहले से चुने गए हैं। तीसरा विद्यार्थी शेष (5) में से \(\binom{5}{1}=5\) तरीकों से चुनेगा।
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(8) विद्यार्थियों में से (4) चुनने हैं और दो विशेष विद्यार्थी साथ-साथ नहीं चुने जाने चाहिए। कितने तरीके हैं?
From (8) students, (4) are to be selected and two special students should not be selected together. How many ways are there?
#combinations
#class11
#easy
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A (55)
B (60)
C (65)
D (70)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{8}{4}=70\), and ways with both special students are \(\binom{6}{2}=15\). Thus (70-15=55) ways remain.
Step 2
Why this answer is correct
The correct answer is A. (55). Total ways are \(\binom{8}{4}=70\), and ways with both special students are \(\binom{6}{2}=15\). Thus (70-15=55) ways remain.
Step 3
Exam Tip
कुल \(\binom{8}{4}=70\) और दोनों विशेष साथ हों तो \(\binom{6}{2}=15\) तरीके हैं। इसलिए (70-15=55) तरीके बचेंगे।
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\(\binom{14}{1}\) का मान क्या है?
What is the value of \(\binom{14}{1}\)?
#combinations
#class11
#easy
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A (1)
B (13)
C (14)
D (28)
Explanation opens after your attempt
Step 1
Concept
\(\binom{n}{1}=n\). Therefore \(\binom{14}{1}=14\).
Step 2
Why this answer is correct
The correct answer is C. (14). \(\binom{n}{1}=n\). Therefore \(\binom{14}{1}=14\).
Step 3
Exam Tip
\(\binom{n}{1}=n\) होता है। इसलिए \(\binom{14}{1}=14\) है।
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\(\binom{15}{0}\) का मान क्या होगा?
What will be the value of \(\binom{15}{0}\)?
#combinations
#class11
#easy
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A (0)
B (1)
C (15)
D (30)
Explanation opens after your attempt
Step 1
Concept
There is one way to choose zero objects. Hence \(\binom{15}{0}=1\).
Step 2
Why this answer is correct
The correct answer is B. (1). There is one way to choose zero objects. Hence \(\binom{15}{0}=1\).
Step 3
Exam Tip
शून्य वस्तु चुनने का एक तरीका होता है। इसलिए \(\binom{15}{0}=1\) है।
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\(\binom{10}{10}\) का मान क्या है?
What is the value of \(\binom{10}{10}\)?
#combinations
#class11
#easy
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A (0)
B (1)
C (10)
D (100)
Explanation opens after your attempt
Step 1
Concept
There is exactly one way to choose all objects. Thus \(\binom{10}{10}=1\).
Step 2
Why this answer is correct
The correct answer is B. (1). There is exactly one way to choose all objects. Thus \(\binom{10}{10}=1\).
Step 3
Exam Tip
सभी वस्तुएं चुनने का एक ही तरीका होता है। अतः \(\binom{10}{10}=1\) है।
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\(\binom{13}{2}\) का मान ज्ञात कीजिए।
Find the value of \(\binom{13}{2}\).
#combinations
#class11
#easy
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A (26)
B (39)
C (78)
D (156)
Explanation opens after your attempt
Step 1
Concept
\(\binom{13}{2}=\frac{13\cdot12}{2}=78\). For choosing two objects, take half of the product.
Step 2
Why this answer is correct
The correct answer is C. (78). \(\binom{13}{2}=\frac{13\cdot12}{2}=78\). For choosing two objects, take half of the product.
Step 3
Exam Tip
\(\binom{13}{2}=\frac{13\cdot12}{2}=78\) है। दो वस्तुओं के चयन में आधा गुणनफल लें।
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\(\binom{12}{4}\) का मान क्या है?
What is the value of \(\binom{12}{4}\)?
#combinations
#class11
#easy
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A (48)
B (220)
C (495)
D (11880)
Explanation opens after your attempt
Step 1
Concept
\(\binom{12}{4}=495\). Dividing by (4!) is necessary in calculation.
Step 2
Why this answer is correct
The correct answer is C. (495). \(\binom{12}{4}=495\). Dividing by (4!) is necessary in calculation.
Step 3
Exam Tip
\(\binom{12}{4}=495\) होता है। गणना में (4!) से भाग देना जरूरी है।
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\(\binom{9}{5}\) किसके बराबर है?
\(\binom{9}{5}\) is equal to which expression?
#combinations
#class11
#easy
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A \(\binom{9}{4}\)
B \(\binom{9}{3}\)
C \(\binom{5}{9}\)
D \(\binom{4}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(\binom{9}{4}\)
Step 1
Concept
Because \(\binom{n}{r}=\binom{n}{n-r}\). Here (9-5=4).
Step 2
Why this answer is correct
The correct answer is A. \(\binom{9}{4}\). Because \(\binom{n}{r}=\binom{n}{n-r}\). Here (9-5=4).
Step 3
Exam Tip
क्योंकि \(\binom{n}{r}=\binom{n}{n-r}\) होता है। यहां (9-5=4) है।
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\(\binom{7}{2}+\binom{7}{5}\) का मान क्या है?
What is the value of \(\binom{7}{2}+\binom{7}{5}\)?
#combinations
#class11
#easy
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A (21)
B (35)
C (42)
D (70)
Explanation opens after your attempt
Step 1
Concept
\(\binom{7}{2}=21\) and \(\binom{7}{5}=21\). Therefore the total is (42).
Step 2
Why this answer is correct
The correct answer is C. (42). \(\binom{7}{2}=21\) and \(\binom{7}{5}=21\). Therefore the total is (42).
Step 3
Exam Tip
\(\binom{7}{2}=21\) और \(\binom{7}{5}=21\) हैं। इसलिए कुल (42) है।
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यदि \(\binom{n}{2}=21\) है तो (n) का मान क्या है?
If \(\binom{n}{2}=21\), what is the value of (n)?
#combinations
#class11
#easy
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A (6)
B (7)
C (8)
D (9)
Explanation opens after your attempt
Step 1
Concept
\(\binom{7}{2}=21\). Hence (n=7) is correct.
Step 2
Why this answer is correct
The correct answer is B. (7). \(\binom{7}{2}=21\). Hence (n=7) is correct.
Step 3
Exam Tip
\(\binom{7}{2}=21\) होता है। इसलिए (n=7) सही है।
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यदि \(\binom{n}{1}+\binom{n}{0}=16\) है तो (n) का मान क्या है?
If \(\binom{n}{1}+\binom{n}{0}=16\), what is the value of (n)?
#combinations
#class11
#easy
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A (14)
B (15)
C (16)
D (17)
Explanation opens after your attempt
Step 1
Concept
It gives (n+1=16). Therefore (n=15).
Step 2
Why this answer is correct
The correct answer is B. (15). It gives (n+1=16). Therefore (n=15).
Step 3
Exam Tip
यह (n+1=16) देता है। इसलिए (n=15) होगा।
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\(\binom{6}{2}+\binom{6}{4}\) का मान ज्ञात कीजिए।
Find the value of \(\binom{6}{2}+\binom{6}{4}\).
#combinations
#class11
#easy
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A (15)
B (24)
C (30)
D (45)
Explanation opens after your attempt
Step 1
Concept
\(\binom{6}{2}=15\) and \(\binom{6}{4}=15\). The total is (30).
Step 2
Why this answer is correct
The correct answer is C. (30). \(\binom{6}{2}=15\) and \(\binom{6}{4}=15\). The total is (30).
Step 3
Exam Tip
\(\binom{6}{2}=15\) और \(\binom{6}{4}=15\) हैं। कुल (30) मिलेगा।
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पास्कल पहचान के अनुसार \(\binom{5}{2}+\binom{5}{3}\) किसके बराबर है?
By Pascal's identity, \(\binom{5}{2}+\binom{5}{3}\) is equal to which expression?
#combinations
#class11
#easy
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A \(\binom{6}{3}\)
B \(\binom{6}{2}\)
C \(\binom{5}{5}\)
D \(\binom{10}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(\binom{6}{3}\)
Step 1
Concept
Pascal's identity is \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{6}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(\binom{6}{3}\). Pascal's identity is \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{6}{3}\).
Step 3
Exam Tip
पास्कल पहचान \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) है। इसलिए उत्तर \(\binom{6}{3}\) है।
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(6) अलग-अलग सिक्कों में से कम से कम (1) सिक्का चुनने के कितने तरीके हैं?
In how many ways can at least (1) coin be selected from (6) different coins?
#combinations
#class11
#easy
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A (31)
B (32)
C (63)
D (64)
Explanation opens after your attempt
Step 1
Concept
The ways for at least (1) selection are \(2^6-1=63\). Do not forget to subtract the empty selection.
Step 2
Why this answer is correct
The correct answer is C. (63). The ways for at least (1) selection are \(2^6-1=63\). Do not forget to subtract the empty selection.
Step 3
Exam Tip
कम से कम (1) चयन के तरीके \(2^6-1=63\) हैं। खाली चयन को घटाना न भूलें।
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(5) अलग-अलग कार्डों में से किसी भी संख्या में कार्ड चुनने के कितने तरीके हैं?
In how many ways can any number of cards be selected from (5) different cards?
#combinations
#class11
#easy
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A (5)
B (25)
C (31)
D (32)
Explanation opens after your attempt
Step 1
Concept
Each card may be selected or not selected. Hence the total ways are \(2^5=32\).
Step 2
Why this answer is correct
The correct answer is D. (32). Each card may be selected or not selected. Hence the total ways are \(2^5=32\).
Step 3
Exam Tip
हर कार्ड चुना या न चुना जा सकता है। इसलिए कुल तरीके \(2^5=32\) हैं।
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(4) फलों में से कम से कम (2) फल चुनने के कितने तरीके हैं?
In how many ways can at least (2) fruits be selected from (4) fruits?
#combinations
#class11
#easy
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A (6)
B (10)
C (11)
D (16)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{4}{2}+\binom{4}{3}+\binom{4}{4}=6+4+1=11\). Add all possible values in at least questions.
Step 2
Why this answer is correct
The correct answer is C. (11). The ways are \(\binom{4}{2}+\binom{4}{3}+\binom{4}{4}=6+4+1=11\). Add all possible values in at least questions.
Step 3
Exam Tip
तरीके \(\binom{4}{2}+\binom{4}{3}+\binom{4}{4}=6+4+1=11\) हैं। कम से कम में सभी मान जोड़ें।
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(7) विषयों में से अधिकतम (1) विषय चुनने के कितने तरीके हैं?
In how many ways can at most (1) subject be selected from (7) subjects?
#combinations
#class11
#easy
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A (7)
B (8)
C (14)
D (49)
Explanation opens after your attempt
Step 1
Concept
At most (1) means selecting (0) or (1). Hence \(\binom{7}{0}+\binom{7}{1}=1+7=8\).
Step 2
Why this answer is correct
The correct answer is B. (8). At most (1) means selecting (0) or (1). Hence \(\binom{7}{0}+\binom{7}{1}=1+7=8\).
Step 3
Exam Tip
अधिकतम (1) का अर्थ (0) या (1) चयन है। इसलिए \(\binom{7}{0}+\binom{7}{1}=1+7=8\) है।
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(9) अंकों में से (2) अंक चुनने हैं ताकि अंकों का क्रम महत्वपूर्ण न हो। कितने तरीके हैं?
From (9) digits, (2) digits are to be chosen and the order is not important. How many ways are there?
#combinations
#class11
#easy
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A (18)
B (36)
C (72)
D (81)
Explanation opens after your attempt
Step 1
Concept
Order is not important, so \(\binom{9}{2}=36\). This is the key idea of combination.
Step 2
Why this answer is correct
The correct answer is B. (36). Order is not important, so \(\binom{9}{2}=36\). This is the key idea of combination.
Step 3
Exam Tip
क्रम महत्वपूर्ण नहीं है इसलिए \(\binom{9}{2}=36\) है। यही संयोजन की पहचान है।
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(10) अक्षरों में से (4) अक्षरों का चयन कितने तरीकों से हो सकता है?
In how many ways can (4) letters be selected from (10) letters?
#combinations
#class11
#easy
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A (40)
B (120)
C (210)
D (5040)
Explanation opens after your attempt
Step 1
Concept
Letters are only being selected, so \(\binom{10}{4}=210\). Use combination when arrangement is not asked.
Step 2
Why this answer is correct
The correct answer is C. (210). Letters are only being selected, so \(\binom{10}{4}=210\). Use combination when arrangement is not asked.
Step 3
Exam Tip
अक्षर केवल चुने जा रहे हैं इसलिए \(\binom{10}{4}=210\) होगा। व्यवस्था न पूछी जाए तो संयोजन लगाएं।
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(6) चित्रों में से (2) चित्र चुनकर पोस्टर में लगाने हैं। कितने चयन संभव हैं?
(2) pictures are to be selected from (6) pictures for a poster. How many selections are possible?
#combinations
#class11
#easy
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A (12)
B (15)
C (30)
D (36)
Explanation opens after your attempt
Step 1
Concept
The ways to choose two pictures are \(\binom{6}{2}=15\). If positions are not given, consider only selection.
Step 2
Why this answer is correct
The correct answer is B. (15). The ways to choose two pictures are \(\binom{6}{2}=15\). If positions are not given, consider only selection.
Step 3
Exam Tip
दो चित्र चुनने के तरीके \(\binom{6}{2}=15\) हैं। लगाने की जगह न दी हो तो केवल चयन मानें।
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(7) स्वादों में से (3) स्वादों वाली आइसक्रीम कितने तरीकों से चुनी जा सकती है?
In how many ways can an ice cream with (3) flavors be selected from (7) flavors?
#combinations
#class11
#easy
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A (21)
B (30)
C (35)
D (210)
Explanation opens after your attempt
Step 1
Concept
The (3) flavors can be selected in \(\binom{7}{3}=35\) ways. The order of flavors is usually not counted.
Step 2
Why this answer is correct
The correct answer is C. (35). The (3) flavors can be selected in \(\binom{7}{3}=35\) ways. The order of flavors is usually not counted.
Step 3
Exam Tip
तीन स्वादों का चयन \(\binom{7}{3}=35\) तरीकों से होगा। स्वादों का क्रम सामान्यतः नहीं गिना जाता।
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(8) अलग-अलग टिकटों में से (5) टिकट चुनने के कितने तरीके हैं?
How many ways are there to choose (5) tickets from (8) different tickets?
#combinations
#class11
#easy
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A (40)
B (56)
C (120)
D (336)
Explanation opens after your attempt
Step 1
Concept
\(\binom{8}{5}=\binom{8}{3}=56\). Choosing by the complementary form is easier.
Step 2
Why this answer is correct
The correct answer is B. (56). \(\binom{8}{5}=\binom{8}{3}=56\). Choosing by the complementary form is easier.
Step 3
Exam Tip
\(\binom{8}{5}=\binom{8}{3}=56\) है। पूरक रूप से चुनना आसान है।
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(3) हिंदी और (4) अंग्रेजी पुस्तकों में से कुल (2) पुस्तकें चुननी हैं जिनमें दोनों हिंदी हों। कितने तरीके हैं?
From (3) Hindi and (4) English books, (2) books are to be selected and both must be Hindi. How many ways are there?
#combinations
#class11
#easy
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A (3)
B (6)
C (12)
D (21)
Explanation opens after your attempt
Step 1
Concept
Both books must be Hindi, so there are \(\binom{3}{2}=3\) ways. Read the condition and choose the correct group.
Step 2
Why this answer is correct
The correct answer is A. (3). Both books must be Hindi, so there are \(\binom{3}{2}=3\) ways. Read the condition and choose the correct group.
Step 3
Exam Tip
दोनों पुस्तकें हिंदी होनी हैं इसलिए \(\binom{3}{2}=3\) तरीके हैं। शर्त पढ़कर सही समूह चुनें।
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(4) वरिष्ठ और (5) कनिष्ठ विद्यार्थियों में से (2) वरिष्ठ और (1) कनिष्ठ कितने तरीकों से चुने जा सकते हैं?
From (4) senior and (5) junior students, in how many ways can (2) senior and (1) junior students be selected?
#combinations
#class11
#easy
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A (11)
B (20)
C (30)
D (60)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{4}{2}\binom{5}{1}=6\cdot5=30\). Selections from different categories are multiplied.
Step 2
Why this answer is correct
The correct answer is C. (30). The ways are \(\binom{4}{2}\binom{5}{1}=6\cdot5=30\). Selections from different categories are multiplied.
Step 3
Exam Tip
तरीके \(\binom{4}{2}\binom{5}{1}=6\cdot5=30\) हैं। अलग श्रेणियों का चयन गुणा से जुड़ता है।
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(5) विज्ञान और (6) कला विद्यार्थियों में से (1) विज्ञान और (3) कला विद्यार्थी कितने तरीकों से चुने जा सकते हैं?
From (5) science and (6) arts students, in how many ways can (1) science and (3) arts students be selected?
#combinations
#class11
#easy
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A (50)
B (75)
C (100)
D (120)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{5}{1}\binom{6}{3}=5\cdot20=100\). Separate the conditions first and then multiply.
Step 2
Why this answer is correct
The correct answer is C. (100). The ways are \(\binom{5}{1}\binom{6}{3}=5\cdot20=100\). Separate the conditions first and then multiply.
Step 3
Exam Tip
तरीके \(\binom{5}{1}\binom{6}{3}=5\cdot20=100\) हैं। पहले शर्तों को अलग करें फिर गुणा करें।
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(6) मिठाइयों में से ठीक (2) मिठाइयां चुनने के कितने तरीके हैं?
How many ways are there to choose exactly (2) sweets from (6) sweets?
#combinations
#class11
#easy
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A (12)
B (15)
C (30)
D (36)
Explanation opens after your attempt
Step 1
Concept
Choosing exactly (2) means \(\binom{6}{2}\). Its value is (15).
Step 2
Why this answer is correct
The correct answer is B. (15). Choosing exactly (2) means \(\binom{6}{2}\). Its value is (15).
Step 3
Exam Tip
ठीक (2) चुनने का अर्थ \(\binom{6}{2}\) है। इसका मान (15) है।
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(9) रंगों में से (2) रंगों का झंडा बनाना है लेकिन रंगों का स्थान तय नहीं है। कितने चयन हैं?
A flag is to be made by choosing (2) colors from (9) colors, but the positions of colors are not fixed. How many selections are there?
#combinations
#class11
#easy
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A (18)
B (36)
C (72)
D (81)
Explanation opens after your attempt
Step 1
Concept
Positions are not fixed, so only colors are selected. The ways are \(\binom{9}{2}=36\).
Step 2
Why this answer is correct
The correct answer is B. (36). Positions are not fixed, so only colors are selected. The ways are \(\binom{9}{2}=36\).
Step 3
Exam Tip
स्थान तय नहीं है इसलिए केवल रंगों का चयन है। तरीके \(\binom{9}{2}=36\) हैं।
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(10) अलग-अलग उपहारों में से (0) उपहार या (1) उपहार चुनने के कुल तरीके कितने हैं?
What is the total number of ways to choose (0) gifts or (1) gift from (10) different gifts?
#combinations
#class11
#easy
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A (10)
B (11)
C (20)
D (100)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{10}{0}+\binom{10}{1}=1+10=11\). Include the (0) selection too.
Step 2
Why this answer is correct
The correct answer is B. (11). Total ways are \(\binom{10}{0}+\binom{10}{1}=1+10=11\). Include the (0) selection too.
Step 3
Exam Tip
कुल तरीके \(\binom{10}{0}+\binom{10}{1}=1+10=11\) हैं। (0) चयन को भी शामिल करें।
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(7) अध्यायों में से पुनरावृत्ति के लिए ठीक (4) अध्याय चुनने के कितने तरीके हैं?
How many ways are there to choose exactly (4) chapters from (7) chapters for revision?
#combinations
#class11
#easy
50 50-50 2 wrong hide
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A (28)
B (35)
C (70)
D (210)
Explanation opens after your attempt
Step 1
Concept
The order of chapters is not asked. Therefore \(\binom{7}{4}=35\).
Step 2
Why this answer is correct
The correct answer is B. (35). The order of chapters is not asked. Therefore \(\binom{7}{4}=35\).
Step 3
Exam Tip
अध्यायों का क्रम नहीं पूछा गया है। इसलिए \(\binom{7}{4}=35\) होगा।
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(11) प्रश्नों में से (2) प्रश्न छोड़ने के कितने तरीके हैं?
In how many ways can (2) questions be left out from (11) questions?
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A (22)
B (55)
C (110)
D (121)
Explanation opens after your attempt
Step 1
Concept
We need to choose (2) questions to leave out. Hence there are \(\binom{11}{2}=55\) ways.
Step 2
Why this answer is correct
The correct answer is B. (55). We need to choose (2) questions to leave out. Hence there are \(\binom{11}{2}=55\) ways.
Step 3
Exam Tip
छोड़ने के लिए (2) प्रश्न चुनने हैं। इसलिए \(\binom{11}{2}=55\) तरीके हैं।
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(12) विद्यार्थियों में से (9) विद्यार्थियों को चुनना (12) में से कितने विद्यार्थियों को न चुनने के बराबर है?
Selecting (9) students from (12) students is equal to not selecting how many students from (12)?
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A (2)
B (3)
C (9)
D (12)
Explanation opens after your attempt
Step 1
Concept
\(\binom{12}{9}=\binom{12}{3}\). Selecting (9) is the same as not selecting (3).
Step 2
Why this answer is correct
The correct answer is B. (3). \(\binom{12}{9}=\binom{12}{3}\). Selecting (9) is the same as not selecting (3).
Step 3
Exam Tip
\(\binom{12}{9}=\binom{12}{3}\) होता है। (9) चुनना (3) न चुनने के बराबर है।
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(5) लड़कों और (3) लड़कियों में से कुल (3) विद्यार्थी चुनने हैं जिनमें कम से कम (1) लड़की हो। कितने तरीके हैं?
From (5) boys and (3) girls, (3) students are to be selected with at least (1) girl. How many ways are there?
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A (25)
B (30)
C (46)
D (56)
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Step 1
Concept
Total ways are \(\binom{8}{3}=56\), and all-boy selections are \(\binom{5}{3}=10\). Therefore (56-10=46) ways.
Step 2
Why this answer is correct
The correct answer is C. (46). Total ways are \(\binom{8}{3}=56\), and all-boy selections are \(\binom{5}{3}=10\). Therefore (56-10=46) ways.
Step 3
Exam Tip
कुल \(\binom{8}{3}=56\) हैं और केवल लड़के \(\binom{5}{3}=10\) हैं। इसलिए (56-10=46) तरीके हैं।
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