(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
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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(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
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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(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
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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(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
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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(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
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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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
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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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
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 \(\binom{n}{2}=45\) है तो (n) का मान क्या है?
If \(\binom{n}{2}=45\), what is the value of (n)?
#combinations
#class11
#easy
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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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
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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\(\binom{12}{2}+\binom{12}{10}\) का मान क्या है?
What is the value of \(\binom{12}{2}+\binom{12}{10}\)?
#combinations
#class11
#easy
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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\(\binom{16}{3}\) का मान ज्ञात कीजिए।
Find the value of \(\binom{16}{3}\).
#combinations
#class11
#easy
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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\(\binom{8}{5}\) किसके बराबर है?
\(\binom{8}{5}\) is equal to which expression?
#combinations
#class11
#easy
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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\(\binom{15}{2}\) का मान क्या होगा?
What will be the value of \(\binom{15}{2}\)?
#combinations
#class11
#easy
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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\(\binom{11}{11}\) का मान ज्ञात कीजिए।
Find the value of \(\binom{11}{11}\).
#combinations
#class11
#easy
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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\(\binom{17}{0}\) का मान क्या है?
What is the value of \(\binom{17}{0}\)?
#combinations
#class11
#easy
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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\(\binom{18}{1}\) का मान क्या है?
What is the value of \(\binom{18}{1}\)?
#combinations
#class11
#easy
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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(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
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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(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
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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(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
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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(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
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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(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
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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(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
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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(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
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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(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
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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(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
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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(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
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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(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
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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(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
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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(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
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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(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
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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