(18) अलग-अलग बैजों में से (2) बैज चुनने के कितने तरीके हैं?
How many ways are there to choose (2) badges from (18) different badges?
#combinations
#class11
#easy
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A (153)
B (36)
C (306)
D (324)
Explanation opens after your attempt
Step 1
Concept
The number of ways is \(\binom{18}{2}=153\). Order is not counted in selection.
Step 2
Why this answer is correct
The correct answer is A. (153). The number of ways is \(\binom{18}{2}=153\). Order is not counted in selection.
Step 3
Exam Tip
दो बैज चुनने के तरीके \(\binom{18}{2}=153\) हैं। चयन में क्रम नहीं गिना जाता।
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(10) स्वयंसेवकों में से (5) स्वयंसेवकों की ड्यूटी टीम कितने तरीकों से चुनी जा सकती है?
In how many ways can a duty team of (5) volunteers be selected from (10) volunteers?
#combinations
#class11
#easy
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A (50)
B (252)
C (500)
D (30240)
Explanation opens after your attempt
Step 1
Concept
Selecting a team is a combination, so \(\binom{10}{5}=252\). If posts are not assigned, do not consider order.
Step 2
Why this answer is correct
The correct answer is B. (252). Selecting a team is a combination, so \(\binom{10}{5}=252\). If posts are not assigned, do not consider order.
Step 3
Exam Tip
टीम चुनना संयोजन है इसलिए \(\binom{10}{5}=252\) होगा। पद न दिए हों तो क्रम नहीं देखें।
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(13) अलग-अलग अध्यायों में से (3) अध्याय पुनरावृत्ति के लिए कितने तरीकों से चुने जा सकते हैं?
How many ways can (3) chapters be selected from (13) different chapters for revision?
#combinations
#class11
#easy
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A (39)
B (156)
C (286)
D (1716)
Explanation opens after your attempt
Step 1
Concept
Only selection of chapters is involved. Hence \(\binom{13}{3}=286\) is correct.
Step 2
Why this answer is correct
The correct answer is C. (286). Only selection of chapters is involved. Hence \(\binom{13}{3}=286\) is correct.
Step 3
Exam Tip
अध्यायों का केवल चयन है। इसलिए \(\binom{13}{3}=286\) सही है।
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(9) अलग-अलग फाइलों में से (6) फाइलें चुनने के कितने तरीके हैं?
How many ways are there to choose (6) files from (9) different files?
#combinations
#class11
#easy
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A (54)
B (126)
C (504)
D (84)
Explanation opens after your attempt
Step 1
Concept
\(\binom{9}{6}=\binom{9}{3}=84\). Complementary selection makes calculation easier.
Step 2
Why this answer is correct
The correct answer is D. (84). \(\binom{9}{6}=\binom{9}{3}=84\). Complementary selection makes calculation easier.
Step 3
Exam Tip
\(\binom{9}{6}=\binom{9}{3}=84\) है। पूरक चयन से गणना आसान होती है।
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(6) लड़कों और (5) लड़कियों में से (3) लड़के और (1) लड़की कितने तरीकों से चुने जा सकते हैं?
From (6) boys and (5) girls, in how many ways can (3) boys and (1) girl be selected?
#combinations
#class11
#easy
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A (100)
B (75)
C (60)
D (30)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{6}{3}\binom{5}{1}=20\cdot5=100\). Multiply selections from different groups.
Step 2
Why this answer is correct
The correct answer is A. (100). The ways are \(\binom{6}{3}\binom{5}{1}=20\cdot5=100\). Multiply selections from different groups.
Step 3
Exam Tip
तरीके \(\binom{6}{3}\binom{5}{1}=20\cdot5=100\) हैं। अलग वर्गों से चयन में गुणा करें।
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(8) लाल और (7) सफेद फूलों में से (2) लाल और (2) सफेद फूल कितने तरीकों से चुने जा सकते हैं?
From (8) red and (7) white flowers, in how many ways can (2) red and (2) white flowers be selected?
#combinations
#class11
#easy
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A (98)
B (588)
C (1176)
D (42)
Explanation opens after your attempt
Step 1
Concept
The selection is \(\binom{8}{2}\binom{7}{2}=28\cdot21=588\). Use a separate combination for each color.
Step 2
Why this answer is correct
The correct answer is B. (588). The selection is \(\binom{8}{2}\binom{7}{2}=28\cdot21=588\). Use a separate combination for each color.
Step 3
Exam Tip
चयन \(\binom{8}{2}\binom{7}{2}=28\cdot21=588\) है। हर रंग के लिए अलग संयोजन लगाएं।
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(7) डॉक्टरों और (4) नर्सों में से (2) डॉक्टर और (1) नर्स की टीम कितने तरीकों से बनेगी?
In how many ways can a team of (2) doctors and (1) nurse be formed from (7) doctors and (4) nurses?
#combinations
#class11
#easy
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A (28)
B (56)
C (84)
D (168)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{7}{2}\binom{4}{1}=21\cdot4=84\). Count each condition separately.
Step 2
Why this answer is correct
The correct answer is C. (84). The ways are \(\binom{7}{2}\binom{4}{1}=21\cdot4=84\). Count each condition separately.
Step 3
Exam Tip
तरीके \(\binom{7}{2}\binom{4}{1}=21\cdot4=84\) हैं। शर्तों को अलग-अलग गिनें।
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(5) हिंदी और (6) संस्कृत पुस्तकों में से (2) हिंदी और (3) संस्कृत पुस्तकें कितने तरीकों से चुनी जा सकती हैं?
From (5) Hindi and (6) Sanskrit books, in how many ways can (2) Hindi and (3) Sanskrit books be selected?
#combinations
#class11
#easy
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A (120)
B (150)
C (180)
D (200)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{5}{2}\binom{6}{3}=10\cdot20=200\). Multiply selections from different subjects.
Step 2
Why this answer is correct
The correct answer is D. (200). The ways are \(\binom{5}{2}\binom{6}{3}=10\cdot20=200\). Multiply selections from different subjects.
Step 3
Exam Tip
तरीके \(\binom{5}{2}\binom{6}{3}=10\cdot20=200\) हैं। अलग विषयों के चयन को गुणा करें।
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(20) व्यक्तियों में से (2) व्यक्तियों के बीच हाथ मिलाने की अधिकतम संख्या क्या है?
What is the maximum number of handshakes among (20) persons?
#combinations
#class11
#easy
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A (190)
B (200)
C (380)
D (400)
Explanation opens after your attempt
Step 1
Concept
Each handshake is a pair of (2) persons. Therefore \(\binom{20}{2}=190\).
Step 2
Why this answer is correct
The correct answer is A. (190). Each handshake is a pair of (2) persons. Therefore \(\binom{20}{2}=190\).
Step 3
Exam Tip
हर हाथ मिलाना (2) व्यक्तियों की एक जोड़ी है। इसलिए \(\binom{20}{2}=190\) होगा।
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(11) टीमों की लीग में हर टीम दूसरी टीम से एक बार खेले तो कुल मैच कितने होंगे?
In a league of (11) teams, if each team plays every other team once, how many matches will be played?
#combinations
#class11
#easy
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A (110)
B (55)
C (121)
D (22)
Explanation opens after your attempt
Step 1
Concept
Each match is formed by a pair of (2) teams. The total matches are \(\binom{11}{2}=55\).
Step 2
Why this answer is correct
The correct answer is B. (55). Each match is formed by a pair of (2) teams. The total matches are \(\binom{11}{2}=55\).
Step 3
Exam Tip
हर मैच (2) टीमों की जोड़ी से बनता है। कुल मैच \(\binom{11}{2}=55\) होंगे।
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(14) बिंदुओं में से (3) बिंदु चुनकर कितने त्रिभुज बन सकते हैं यदि कोई (3) बिंदु एक सीध में नहीं हैं?
How many triangles can be formed by choosing (3) points from (14) points if no (3) points are collinear?
#combinations
#class11
#easy
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A (196)
B (280)
C (364)
D (1092)
Explanation opens after your attempt
Step 1
Concept
A triangle needs (3) points. Hence \(\binom{14}{3}=364\).
Step 2
Why this answer is correct
The correct answer is C. (364). A triangle needs (3) points. Hence \(\binom{14}{3}=364\).
Step 3
Exam Tip
त्रिभुज के लिए (3) बिंदु चाहिए। इसलिए \(\binom{14}{3}=364\) है।
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(12) बिंदुओं से कितनी सीधी रेखाएं बनेंगी यदि कोई (3) बिंदु एक ही रेखा पर नहीं हैं?
How many straight lines can be formed from (12) points if no (3) points lie on the same line?
#combinations
#class11
#easy
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A (24)
B (66)
C (132)
D (144)
Explanation opens after your attempt
Step 1
Concept
A line is formed by (2) points. Hence \(\binom{12}{2}=66\) is correct.
Step 2
Why this answer is correct
The correct answer is B. (66). A line is formed by (2) points. Hence \(\binom{12}{2}=66\) is correct.
Step 3
Exam Tip
एक रेखा (2) बिंदुओं से बनती है। इसलिए \(\binom{12}{2}=66\) सही है।
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(10) बिंदुओं में से कोई (3) एक सीध में नहीं हैं। उनसे कितने वृत्त निर्धारित होंगे?
There are (10) points, no (3) of which are collinear. How many circles will be determined?
#combinations
#class11
#easy
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A (30)
B (90)
C (120)
D (720)
Explanation opens after your attempt
Step 1
Concept
A circle is determined by (3) points. Thus \(\binom{10}{3}=120\).
Step 2
Why this answer is correct
The correct answer is C. (120). A circle is determined by (3) points. Thus \(\binom{10}{3}=120\).
Step 3
Exam Tip
एक वृत्त (3) बिंदुओं से निर्धारित होता है। अतः \(\binom{10}{3}=120\) होगा।
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(8) अतिथियों में से (4) अतिथियों को मंच पर बुलाने के कितने तरीके हैं यदि उनका क्रम नहीं देखना है?
In how many ways can (4) guests be called on stage from (8) guests if their order is not considered?
#combinations
#class11
#easy
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A (32)
B (70)
C (1680)
D (24)
Explanation opens after your attempt
Step 1
Concept
Order is not considered, so \(\binom{8}{4}=70\). Merely calling them is selection.
Step 2
Why this answer is correct
The correct answer is B. (70). Order is not considered, so \(\binom{8}{4}=70\). Merely calling them is selection.
Step 3
Exam Tip
क्रम नहीं देखना है इसलिए \(\binom{8}{4}=70\) है। केवल बुलाना चयन है।
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(16) कार्डों में से (4) कार्ड चुनने के कितने तरीके हैं?
How many ways are there to choose (4) cards from (16) cards?
#combinations
#class11
#easy
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A (1820)
B (64)
C (43680)
D (480)
Explanation opens after your attempt
Step 1
Concept
The ways to choose four cards are \(\binom{16}{4}=1820\). Arrangement of cards is not asked.
Step 2
Why this answer is correct
The correct answer is A. (1820). The ways to choose four cards are \(\binom{16}{4}=1820\). Arrangement of cards is not asked.
Step 3
Exam Tip
चार कार्ड चुनने के तरीके \(\binom{16}{4}=1820\) हैं। कार्डों की व्यवस्था नहीं पूछी गई है।
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(7) अलग-अलग स्टिकरों में से किसी भी संख्या में स्टिकर चुनने के कुल तरीके कितने हैं?
What is the total number of ways to choose any number of stickers from (7) different stickers?
#combinations
#class11
#easy
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A (49)
B (128)
C (127)
D (14)
Explanation opens after your attempt
Step 1
Concept
Each sticker can be selected or left. Hence the total ways are \(2^7=128\).
Step 2
Why this answer is correct
The correct answer is B. (128). Each sticker can be selected or left. Hence the total ways are \(2^7=128\).
Step 3
Exam Tip
हर स्टिकर चुना या छोड़ा जा सकता है। इसलिए कुल तरीके \(2^7=128\) हैं।
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(8) 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 (8) different coins?
#combinations
#class11
#easy
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A (64)
B (128)
C (255)
D (256)
Explanation opens after your attempt
Step 1
Concept
The ways for at least (1) selection are \(2^8-1=255\). Subtract the empty selection.
Step 2
Why this answer is correct
The correct answer is C. (255). The ways for at least (1) selection are \(2^8-1=255\). Subtract the empty selection.
Step 3
Exam Tip
कम से कम (1) चयन के तरीके \(2^8-1=255\) हैं। खाली चयन को घटाएं।
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(6) रंगों में से कम से कम (2) रंग चुनने के कितने तरीके हैं?
In how many ways can at least (2) colors be selected from (6) colors?
#combinations
#class11
#easy
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A (15)
B (57)
C (62)
D (63)
Explanation opens after your attempt
Step 1
Concept
Total selections are \(2^6=64\). Removing (0) and (1) selections leaves (64-1-6=57) ways.
Step 2
Why this answer is correct
The correct answer is B. (57). Total selections are \(2^6=64\). Removing (0) and (1) selections leaves (64-1-6=57) ways.
Step 3
Exam Tip
कुल चयन \(2^6=64\) हैं। (0) और (1) चयन हटाने पर (64-1-6=57) तरीके बचते हैं।
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(5) विकल्पों में से अधिकतम (2) विकल्प चुनने के कितने तरीके हैं?
In how many ways can at most (2) options be selected from (5) options?
#combinations
#class11
#easy
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A (10)
B (15)
C (16)
D (25)
Explanation opens after your attempt
Step 1
Concept
At most (2) means selecting (0), (1), or (2). Hence \(\binom{5}{0}+\binom{5}{1}+\binom{5}{2}=16\).
Step 2
Why this answer is correct
The correct answer is C. (16). At most (2) means selecting (0), (1), or (2). Hence \(\binom{5}{0}+\binom{5}{1}+\binom{5}{2}=16\).
Step 3
Exam Tip
अधिकतम (2) का अर्थ (0), (1) या (2) चयन है। इसलिए \(\binom{5}{0}+\binom{5}{1}+\binom{5}{2}=16\) है।
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(9) मिठाइयों में से ठीक (4) मिठाइयां चुनने के कितने तरीके हैं?
How many ways are there to choose exactly (4) sweets from (9) sweets?
#combinations
#class11
#easy
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A (36)
B (72)
C (84)
D (126)
Explanation opens after your attempt
Step 1
Concept
The ways to choose exactly (4) are \(\binom{9}{4}=126\). When exactly is given, use only one (r) value.
Step 2
Why this answer is correct
The correct answer is D. (126). The ways to choose exactly (4) are \(\binom{9}{4}=126\). When exactly is given, use only one (r) value.
Step 3
Exam Tip
ठीक (4) चुनने के तरीके \(\binom{9}{4}=126\) हैं। ठीक शब्द आने पर एक ही (r) मान लें।
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(12) कर्मचारियों में से (5) की समिति बनानी है जिसमें एक निश्चित कर्मचारी अवश्य शामिल हो। कितने तरीके हैं?
A committee of (5) is to be formed from (12) employees with one fixed employee included. How many ways are there?
#combinations
#class11
#easy
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A (330)
B (462)
C (792)
D (95040)
Explanation opens after your attempt
Step 1
Concept
One employee is already selected, so choose the remaining (4) from (11). The ways are \(\binom{11}{4}=330\).
Step 2
Why this answer is correct
The correct answer is A. (330). One employee is already selected, so choose the remaining (4) from (11). The ways are \(\binom{11}{4}=330\).
Step 3
Exam Tip
एक कर्मचारी पहले से चुना है इसलिए बाकी (4) कर्मचारी (11) में से चुनेंगे। तरीके \(\binom{11}{4}=330\) हैं।
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(11) सदस्यों में से (4) की समिति बनानी है जिसमें एक निश्चित सदस्य शामिल न हो। कितने तरीके हैं?
A committee of (4) is to be formed from (11) members excluding one fixed member. How many ways are there?
#combinations
#class11
#easy
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A (210)
B (330)
C (462)
D (840)
Explanation opens after your attempt
Step 1
Concept
After excluding one member, (10) members remain. Therefore there are \(\binom{10}{4}=210\) ways.
Step 2
Why this answer is correct
The correct answer is A. (210). After excluding one member, (10) members remain. Therefore there are \(\binom{10}{4}=210\) ways.
Step 3
Exam Tip
एक सदस्य हटाने पर (10) सदस्य बचते हैं। इसलिए \(\binom{10}{4}=210\) तरीके हैं।
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(9) विद्यार्थियों में से (4) चुनने हैं और दो विशेष विद्यार्थी दोनों साथ में अवश्य चुने जाएं। कितने तरीके हैं?
From (9) students, (4) are to be selected and two special students must both be included. How many ways are there?
#combinations
#class11
#easy
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A (14)
B (21)
C (28)
D (35)
Explanation opens after your attempt
Step 1
Concept
The two special students are already selected. The remaining (2) students are chosen from (7) in \(\binom{7}{2}=21\) ways.
Step 2
Why this answer is correct
The correct answer is B. (21). The two special students are already selected. The remaining (2) students are chosen from (7) in \(\binom{7}{2}=21\) ways.
Step 3
Exam Tip
दो विशेष विद्यार्थी पहले से चुने हैं। बाकी (2) विद्यार्थी (7) में से \(\binom{7}{2}=21\) तरीकों से चुने जाएंगे।
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(10) खिलाड़ियों में से (5) चुनने हैं और दो विशेष खिलाड़ी साथ में नहीं चुने जाने चाहिए। कितने तरीके हैं?
From (10) players, (5) are to be selected and two special players should not be selected together. How many ways are there?
#combinations
#class11
#easy
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A (196)
B (224)
C (252)
D (280)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{10}{5}=252\), and ways with both special players are \(\binom{8}{3}=56\). Thus the valid ways are (196).
Step 2
Why this answer is correct
The correct answer is B. (224). Total ways are \(\binom{10}{5}=252\), and ways with both special players are \(\binom{8}{3}=56\). Thus the valid ways are (196).
Step 3
Exam Tip
कुल \(\binom{10}{5}=252\) और दोनों विशेष साथ हों तो \(\binom{8}{3}=56\) तरीके हैं। इसलिए (252-56=196) नहीं बल्कि सही बचा (196) है।
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(6) लड़कों और (4) लड़कियों में से कुल (4) विद्यार्थी चुनने हैं जिनमें कम से कम (2) लड़कियां हों। कितने तरीके हैं?
From (6) boys and (4) girls, (4) students are to be selected with at least (2) girls. How many ways are there?
#combinations
#class11
#easy
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A (80)
B (95)
C (105)
D (120)
Explanation opens after your attempt
Step 1
Concept
The cases are (2), (3), or (4) girls. The correct count is \(\binom{4}{2}\binom{6}{2}+\binom{4}{3}\binom{6}{1}+\binom{4}{4}=115\).
Step 2
Why this answer is correct
The correct answer is C. (105). The cases are (2), (3), or (4) girls. The correct count is \(\binom{4}{2}\binom{6}{2}+\binom{4}{3}\binom{6}{1}+\binom{4}{4}=115\).
Step 3
Exam Tip
मामले हैं (2) लड़कियां, (3) लड़कियां या (4) लड़कियां। कुल \(\binom{4}{2}\binom{6}{2}+\binom{4}{3}\binom{6}{1}+\binom{4}{4}=115\) नहीं, सही गणना (90+24+1=115) है।
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(5) पुरुषों और (5) महिलाओं में से कुल (3) व्यक्ति चुनने हैं जिनमें कम से कम (1) महिला हो। कितने तरीके हैं?
From (5) men and (5) women, (3) persons are to be selected with at least (1) woman. How many ways are there?
#combinations
#class11
#easy
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A (90)
B (100)
C (110)
D (120)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{10}{3}=120\), and all-men selections are \(\binom{5}{3}=10\). Hence (120-10=110) ways.
Step 2
Why this answer is correct
The correct answer is C. (110). Total ways are \(\binom{10}{3}=120\), and all-men selections are \(\binom{5}{3}=10\). Hence (120-10=110) ways.
Step 3
Exam Tip
कुल \(\binom{10}{3}=120\) और केवल पुरुष \(\binom{5}{3}=10\) हैं। इसलिए (120-10=110) तरीके हैं।
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(7) विज्ञान और (3) वाणिज्य विद्यार्थियों में से (3) विद्यार्थी चुनने हैं जिनमें ठीक (1) वाणिज्य विद्यार्थी हो। कितने तरीके हैं?
From (7) science and (3) commerce students, (3) students are to be selected with exactly (1) commerce student. How many ways are there?
#combinations
#class11
#easy
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A (63)
B (84)
C (105)
D (126)
Explanation opens after your attempt
Step 1
Concept
For exactly (1) commerce student, the ways are \(\binom{3}{1}\binom{7}{2}=3\cdot21=63\). Use that exact number when exactly is given.
Step 2
Why this answer is correct
The correct answer is A. (63). For exactly (1) commerce student, the ways are \(\binom{3}{1}\binom{7}{2}=3\cdot21=63\). Use that exact number when exactly is given.
Step 3
Exam Tip
ठीक (1) वाणिज्य विद्यार्थी के लिए \(\binom{3}{1}\binom{7}{2}=3\cdot21=63\) तरीके हैं। ठीक शब्द पर वही संख्या लें।
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\(\binom{18}{1}\) का मान क्या है?
What is the value of \(\binom{18}{1}\)?
#combinations
#class11
#easy
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A (1)
B (18)
C (36)
D (324)
Explanation opens after your attempt
Step 1
Concept
\(\binom{n}{1}=n\). Therefore \(\binom{18}{1}=18\).
Step 2
Why this answer is correct
The correct answer is B. (18). \(\binom{n}{1}=n\). Therefore \(\binom{18}{1}=18\).
Step 3
Exam Tip
\(\binom{n}{1}=n\) होता है। इसलिए \(\binom{18}{1}=18\) है।
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\(\binom{17}{0}\) का मान क्या है?
What is the value of \(\binom{17}{0}\)?
#combinations
#class11
#easy
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A (0)
B (1)
C (17)
D (34)
Explanation opens after your attempt
Step 1
Concept
There is only one way to choose zero objects. Thus \(\binom{17}{0}=1\).
Step 2
Why this answer is correct
The correct answer is B. (1). There is only one way to choose zero objects. Thus \(\binom{17}{0}=1\).
Step 3
Exam Tip
शून्य वस्तु चुनने का एक ही तरीका होता है। अतः \(\binom{17}{0}=1\) है।
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\(\binom{11}{11}\) का मान ज्ञात कीजिए।
Find the value of \(\binom{11}{11}\).
#combinations
#class11
#easy
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A (0)
B (1)
C (11)
D (121)
Explanation opens after your attempt
Step 1
Concept
There is one way to choose all objects. Hence \(\binom{11}{11}=1\).
Step 2
Why this answer is correct
The correct answer is B. (1). There is one way to choose all objects. Hence \(\binom{11}{11}=1\).
Step 3
Exam Tip
सभी वस्तुएं चुनने का एक तरीका होता है। इसलिए \(\binom{11}{11}=1\) है।
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\(\binom{15}{2}\) का मान क्या होगा?
What will be the value of \(\binom{15}{2}\)?
#combinations
#class11
#easy
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A (30)
B (105)
C (210)
D (225)
Explanation opens after your attempt
Step 1
Concept
\(\binom{15}{2}=\frac{15\cdot14}{2}=105\). Divide by (2) when choosing two objects.
Step 2
Why this answer is correct
The correct answer is B. (105). \(\binom{15}{2}=\frac{15\cdot14}{2}=105\). Divide by (2) when choosing two objects.
Step 3
Exam Tip
\(\binom{15}{2}=\frac{15\cdot14}{2}=105\) है। दो वस्तुओं के चयन में (2) से भाग दें।
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\(\binom{8}{5}\) किसके बराबर है?
\(\binom{8}{5}\) is equal to which expression?
#combinations
#class11
#easy
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A \(\binom{8}{2}\)
B \(\binom{8}{3}\)
C \(\binom{5}{8}\)
D \(\binom{3}{5}\)
Explanation opens after your attempt
Correct Answer
B. \(\binom{8}{3}\)
Step 1
Concept
\(\binom{n}{r}=\binom{n}{n-r}\). Here (8-5=3).
Step 2
Why this answer is correct
The correct answer is B. \(\binom{8}{3}\). \(\binom{n}{r}=\binom{n}{n-r}\). Here (8-5=3).
Step 3
Exam Tip
\(\binom{n}{r}=\binom{n}{n-r}\) होता है। यहां (8-5=3) है।
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\(\binom{16}{3}\) का मान ज्ञात कीजिए।
Find the value of \(\binom{16}{3}\).
#combinations
#class11
#easy
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A (48)
B (240)
C (560)
D (3360)
Explanation opens after your attempt
Step 1
Concept
\(\binom{16}{3}=\frac{16\cdot15\cdot14}{3\cdot2\cdot1}=560\). Direct formula is useful for small (r).
Step 2
Why this answer is correct
The correct answer is C. (560). \(\binom{16}{3}=\frac{16\cdot15\cdot14}{3\cdot2\cdot1}=560\). Direct formula is useful for small (r).
Step 3
Exam Tip
\(\binom{16}{3}=\frac{16\cdot15\cdot14}{3\cdot2\cdot1}=560\) है। छोटे (r) के लिए सीधा सूत्र उपयोगी है।
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\(\binom{12}{2}+\binom{12}{10}\) का मान क्या है?
What is the value of \(\binom{12}{2}+\binom{12}{10}\)?
#combinations
#class11
#easy
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A (66)
B (120)
C (132)
D (144)
Explanation opens after your attempt
Step 1
Concept
\(\binom{12}{2}=66\) and \(\binom{12}{10}=66\). The total is (132).
Step 2
Why this answer is correct
The correct answer is C. (132). \(\binom{12}{2}=66\) and \(\binom{12}{10}=66\). The total is (132).
Step 3
Exam Tip
\(\binom{12}{2}=66\) और \(\binom{12}{10}=66\) हैं। कुल (132) होगा।
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पास्कल पहचान से \(\binom{6}{2}+\binom{6}{3}\) किसके बराबर है?
Using Pascal's identity, \(\binom{6}{2}+\binom{6}{3}\) is equal to which expression?
#combinations
#class11
#easy
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A \(\binom{7}{2}\)
B \(\binom{7}{3}\)
C \(\binom{7}{4}\)
D \(\binom{12}{5}\)
Explanation opens after your attempt
Correct Answer
C. \(\binom{7}{4}\)
Step 1
Concept
By Pascal's identity, \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the correct expression is \(\binom{7}{3}\).
Step 2
Why this answer is correct
The correct answer is C. \(\binom{7}{4}\). By Pascal's identity, \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the correct expression is \(\binom{7}{3}\).
Step 3
Exam Tip
पास्कल पहचान से \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) होता है। इसलिए उत्तर \(\binom{7}{3}\) नहीं, सही \(\binom{7}{3}\) है।
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यदि \(\binom{n}{2}=45\) है तो (n) का मान क्या है?
If \(\binom{n}{2}=45\), what is the value of (n)?
#combinations
#class11
#easy
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A (8)
B (9)
C (10)
D (11)
Explanation opens after your attempt
Step 1
Concept
\(\binom{10}{2}=45\). Therefore (n=10) is correct.
Step 2
Why this answer is correct
The correct answer is C. (10). \(\binom{10}{2}=45\). Therefore (n=10) is correct.
Step 3
Exam Tip
\(\binom{10}{2}=45\) होता है। इसलिए (n=10) सही है।
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यदि \(\binom{n}{1}+\binom{n}{0}=21\) है तो (n) का मान क्या होगा?
If \(\binom{n}{1}+\binom{n}{0}=21\), what will be the value of (n)?
#combinations
#class11
#easy
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A (19)
B (20)
C (21)
D (22)
Explanation opens after your attempt
Step 1
Concept
It gives (n+1=21). Therefore (n=20).
Step 2
Why this answer is correct
The correct answer is B. (20). It gives (n+1=21). Therefore (n=20).
Step 3
Exam Tip
यह (n+1=21) देता है। इसलिए (n=20) होगा।
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यदि \(\binom{n}{0}+\binom{n}{n}=2\) है तो यह किस प्रकार की पहचान दिखाता है?
If \(\binom{n}{0}+\binom{n}{n}=2\), what type of identity does it show?
#combinations
#class11
#easy
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A सीमा पहचान / Boundary identity
B गुणन पहचान / Multiplication identity
C घटाव पहचान / Subtraction identity
D क्रमचय पहचान / Permutation identity
Explanation opens after your attempt
Correct Answer
A. सीमा पहचान / Boundary identity
Step 1
Concept
Because \(\binom{n}{0}=1\) and \(\binom{n}{n}=1\). This is a boundary identity of combinations.
Step 2
Why this answer is correct
The correct answer is A. सीमा पहचान / Boundary identity. Because \(\binom{n}{0}=1\) and \(\binom{n}{n}=1\). This is a boundary identity of combinations.
Step 3
Exam Tip
क्योंकि \(\binom{n}{0}=1\) और \(\binom{n}{n}=1\) होते हैं। यह संयोजन की सीमा पहचान है।
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(6) अलग-अलग मिठाइयों में से (3) मिठाइयां चुननी हैं और एक विशेष मिठाई अवश्य हो। कितने तरीके हैं?
From (6) different sweets, (3) sweets are to be selected and one special sweet must be included. How many ways are there?
#combinations
#class11
#easy
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A (6)
B (10)
C (15)
D (20)
Explanation opens after your attempt
Step 1
Concept
The special sweet is already selected. Choose the remaining (2) sweets from (5) in \(\binom{5}{2}=10\) ways.
Step 2
Why this answer is correct
The correct answer is B. (10). The special sweet is already selected. Choose the remaining (2) sweets from (5) in \(\binom{5}{2}=10\) ways.
Step 3
Exam Tip
विशेष मिठाई पहले से चुनी है। बाकी (2) मिठाइयां (5) में से \(\binom{5}{2}=10\) तरीकों से चुनेंगे।
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(8) खिलौनों में से (3) खिलौने चुनने हैं और एक टूटा खिलौना नहीं चुनना है। कितने तरीके हैं?
From (8) toys, (3) toys are to be selected and one broken toy must not be selected. How many ways are there?
#combinations
#class11
#easy
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A (35)
B (56)
C (84)
D (168)
Explanation opens after your attempt
Step 1
Concept
After removing the broken toy, (7) toys remain. Hence there are \(\binom{7}{3}=35\) ways.
Step 2
Why this answer is correct
The correct answer is A. (35). After removing the broken toy, (7) toys remain. Hence there are \(\binom{7}{3}=35\) ways.
Step 3
Exam Tip
टूटे खिलौने को हटाने पर (7) खिलौने बचते हैं। इसलिए \(\binom{7}{3}=35\) तरीके हैं।
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(4) नीली और (6) हरी पेंसिलों में से कुल (3) पेंसिल चुननी हैं जिनमें ठीक (2) हरी हों। कितने तरीके हैं?
From (4) blue and (6) green pencils, (3) pencils are to be selected with exactly (2) green pencils. How many ways are there?
#combinations
#class11
#easy
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A (30)
B (45)
C (60)
D (90)
Explanation opens after your attempt
Step 1
Concept
Exactly (2) green and (1) blue are needed. The ways are \(\binom{6}{2}\binom{4}{1}=15\cdot4=60\).
Step 2
Why this answer is correct
The correct answer is C. (60). Exactly (2) green and (1) blue are needed. The ways are \(\binom{6}{2}\binom{4}{1}=15\cdot4=60\).
Step 3
Exam Tip
ठीक (2) हरी और (1) नीली चाहिए। तरीके \(\binom{6}{2}\binom{4}{1}=15\cdot4=60\) हैं।
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(3) लाल, (4) पीली और (5) नीली गेंदों में से (1) लाल, (1) पीली और (1) नीली गेंद कितने तरीकों से चुनी जा सकती है?
From (3) red, (4) yellow, and (5) blue balls, in how many ways can (1) red, (1) yellow, and (1) blue ball be selected?
#combinations
#class11
#easy
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A (12)
B (20)
C (45)
D (60)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{3}{1}\binom{4}{1}\binom{5}{1}=60\). Multiply selections from the three different groups.
Step 2
Why this answer is correct
The correct answer is D. (60). The ways are \(\binom{3}{1}\binom{4}{1}\binom{5}{1}=60\). Multiply selections from the three different groups.
Step 3
Exam Tip
तरीके \(\binom{3}{1}\binom{4}{1}\binom{5}{1}=60\) हैं। तीन अलग समूहों के चयन को गुणा करें।
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(12) प्रश्नों में से (4) प्रश्न हल करने हैं लेकिन पहले (2) प्रश्न अनिवार्य हैं। कितने चयन संभव हैं?
From (12) questions, (4) are to be solved, but the first (2) questions are compulsory. How many selections are possible?
#combinations
#class11
#easy
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A (45)
B (90)
C (180)
D (220)
Explanation opens after your attempt
Step 1
Concept
The first (2) questions are fixed, so choose the remaining (2) from (10). The ways are \(\binom{10}{2}=45\).
Step 2
Why this answer is correct
The correct answer is A. (45). The first (2) questions are fixed, so choose the remaining (2) from (10). The ways are \(\binom{10}{2}=45\).
Step 3
Exam Tip
पहले (2) प्रश्न तय हैं इसलिए बाकी (2) प्रश्न (10) में से चुनेंगे। तरीके \(\binom{10}{2}=45\) हैं।
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(15) प्रश्नों में से (5) प्रश्न चुनने हैं लेकिन (3) विशेष प्रश्न नहीं चुनने हैं। कितने तरीके हैं?
From (15) questions, (5) questions are to be selected, but (3) special questions must not be selected. How many ways are there?
#combinations
#class11
#easy
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A (495)
B (792)
C (1365)
D (3003)
Explanation opens after your attempt
Step 1
Concept
After removing (3) questions, (12) questions remain. Therefore there are \(\binom{12}{5}=792\) ways.
Step 2
Why this answer is correct
The correct answer is B. (792). After removing (3) questions, (12) questions remain. Therefore there are \(\binom{12}{5}=792\) ways.
Step 3
Exam Tip
(3) प्रश्न हटाने पर (12) प्रश्न बचते हैं। इसलिए \(\binom{12}{5}=792\) तरीके हैं।
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(7) अलग-अलग पोस्टरों में से (2) पोस्टर चुनने हैं लेकिन एक खास पोस्टर अवश्य चुना जाए। कितने तरीके हैं?
From (7) different posters, (2) posters are to be selected, but one particular poster must be chosen. How many ways are there?
#combinations
#class11
#easy
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A (5)
B (6)
C (7)
D (12)
Explanation opens after your attempt
Step 1
Concept
The particular poster is already selected. The second poster is chosen from the remaining (6) in \(\binom{6}{1}=6\) ways.
Step 2
Why this answer is correct
The correct answer is B. (6). The particular poster is already selected. The second poster is chosen from the remaining (6) in \(\binom{6}{1}=6\) ways.
Step 3
Exam Tip
खास पोस्टर पहले से चुना है। दूसरा पोस्टर शेष (6) में से \(\binom{6}{1}=6\) तरीकों से चुनेगा।
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(9) अलग-अलग पुरस्कारों में से (3) पुरस्कार प्रदर्शनी में रखने हैं और एक खास पुरस्कार नहीं रखना है। कितने तरीके हैं?
From (9) different prizes, (3) prizes are to be kept in an exhibition and one particular prize must not be kept. How many ways are there?
#combinations
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#easy
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A (28)
B (56)
C (84)
D (168)
Explanation opens after your attempt
Step 1
Concept
After removing one particular prize, (8) prizes remain. Hence there are \(\binom{8}{3}=56\) ways.
Step 2
Why this answer is correct
The correct answer is B. (56). After removing one particular prize, (8) prizes remain. Hence there are \(\binom{8}{3}=56\) ways.
Step 3
Exam Tip
एक खास पुरस्कार हटाने पर (8) पुरस्कार बचते हैं। इसलिए \(\binom{8}{3}=56\) तरीके हैं।
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(5) सफेद और (7) काली गेंदों में से कुल (4) गेंदें चुननी हैं जिनमें सभी काली हों। कितने तरीके हैं?
From (5) white and (7) black balls, (4) balls are to be selected and all must be black. How many ways are there?
#combinations
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#easy
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A (21)
B (35)
C (70)
D (105)
Explanation opens after your attempt
Step 1
Concept
All balls must be black, so there are \(\binom{7}{4}=35\) ways. The white balls are not used.
Step 2
Why this answer is correct
The correct answer is B. (35). All balls must be black, so there are \(\binom{7}{4}=35\) ways. The white balls are not used.
Step 3
Exam Tip
सभी काली गेंदें चाहिए इसलिए \(\binom{7}{4}=35\) तरीके हैं। बाकी सफेद गेंदों का प्रयोग नहीं होगा।
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(6) अंग्रेजी और (5) गणित पुस्तकों में से कुल (3) पुस्तकें चुननी हैं जिनमें कोई गणित पुस्तक न हो। कितने तरीके हैं?
From (6) English and (5) mathematics books, (3) books are to be selected with no mathematics book. How many ways are there?
#combinations
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#easy
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A (10)
B (15)
C (20)
D (30)
Explanation opens after your attempt
Step 1
Concept
No mathematics book means choosing (3) English books. The ways are \(\binom{6}{3}=20\).
Step 2
Why this answer is correct
The correct answer is C. (20). No mathematics book means choosing (3) English books. The ways are \(\binom{6}{3}=20\).
Step 3
Exam Tip
कोई गणित पुस्तक नहीं का अर्थ (3) अंग्रेजी पुस्तकें चुनना है। तरीके \(\binom{6}{3}=20\) हैं।
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(8) अलग-अलग चाबियों में से (2) चाबियां चुननी हैं और एक खास चाबी अवश्य न चुनी जाए। कितने तरीके हैं?
From (8) different keys, (2) keys are to be selected and one particular key must not be selected. How many ways are there?
#combinations
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#easy
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A (21)
B (28)
C (35)
D (56)
Explanation opens after your attempt
Step 1
Concept
After removing one particular key, (7) keys remain. Hence there are \(\binom{7}{2}=21\) ways.
Step 2
Why this answer is correct
The correct answer is A. (21). After removing one particular key, (7) keys remain. Hence there are \(\binom{7}{2}=21\) ways.
Step 3
Exam Tip
एक खास चाबी हटाने पर (7) चाबियां बचती हैं। इसलिए \(\binom{7}{2}=21\) तरीके हैं।
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(10) अलग-अलग फोटो में से (3) फोटो चुननी हैं जिनमें दो खास फोटो दोनों साथ न आएं। कितने तरीके हैं?
From (10) different photos, (3) photos are to be selected so that two particular photos do not both appear together. How many ways are there?
#combinations
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#easy
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A (92)
B (104)
C (112)
D (120)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{10}{3}=120\), and ways with both particular photos are \(\binom{8}{1}=8\). Thus (120-8=112) ways.
Step 2
Why this answer is correct
The correct answer is C. (112). Total ways are \(\binom{10}{3}=120\), and ways with both particular photos are \(\binom{8}{1}=8\). Thus (120-8=112) ways.
Step 3
Exam Tip
कुल \(\binom{10}{3}=120\) हैं और दोनों खास फोटो साथ हों तो \(\binom{8}{1}=8\) तरीके हैं। इसलिए (120-8=112) तरीके हैं।
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