Level 61 • 50/50 questions • 40 seconds per question.
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If (2) books are to be selected from (5) books, how many selections are possible?
B. (10)
यह \(\binom{5}{2}=10\) का सरल प्रयोग है। परीक्षा में चयन का क्रम नहीं गिना जाता। / This is a direct use of \(\binom{5}{2}=10\). In exams, order is not counted in selection.
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In how many ways can a team of (3) students be selected from (6) students?
C. (20)
टीम चुनने में क्रम महत्वपूर्ण नहीं होता। इसलिए \(\binom{6}{3}=20\) सही है। / Order is not important while selecting a team. Hence \(\binom{6}{3}=20\) is correct.
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How many ways are there to choose (2) flowers from (7) different flowers?
B. (21)
यह \(\binom{7}{2}=\frac{7\cdot6}{2}=21\) है। ऐसे प्रश्नों में वस्तुओं की जोड़ी बनती है। / It is \(\binom{7}{2}=\frac{7\cdot6}{2}=21\). Such questions form pairs of objects.
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In how many ways can a group of (4) players be formed from (8) players?
C. (70)
समूह में क्रम नहीं होता इसलिए \(\binom{8}{4}=70\) होगा। परीक्षा में समूह शब्द दिखे तो संयोजन लगाएं। / A group has no order, so \(\binom{8}{4}=70\). In exams, the word group usually indicates combination.
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How many ways are there to select (1) question from (9) questions?
C. (9)
एक वस्तु चुनने पर \(\binom{9}{1}=9\) होता है। \(\binom{n}{1}=n\) याद रखें। / Selecting one object gives \(\binom{9}{1}=9\). Remember \(\binom{n}{1}=n\).
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Why is selecting (0) pens from (10) pens considered one way?
A. क्योंकि \(\binom{10}{0}=1\)Because \(\binom{10}{0}=1\)
कुछ भी न चुनना भी एक निश्चित चयन है। इसलिए \(\binom{n}{0}=1\) होता है। / Choosing nothing is also one definite selection. Therefore \(\binom{n}{0}=1\).
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How many ways are there to choose all (5) colors from (5) colors?
B. (1)
सभी वस्तुएं चुनने का केवल एक तरीका होता है। इसलिए \(\binom{5}{5}=1\) है। / There is only one way to choose all objects. Hence \(\binom{5}{5}=1\).
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In how many ways can a pair of friends be selected from (6) friends?
B. (15)
जोड़ी के लिए \(\binom{6}{2}=15\) होगा। जोड़ी में पहला और दूसरा अलग नहीं गिने जाते। / For a pair, \(\binom{6}{2}=15\). In a pair, first and second are not counted separately.
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From (4) boys and (3) girls, in how many ways can (1) boy and (1) girl be selected?
B. (12)
चयन \(\binom{4}{1}\binom{3}{1}=12\) है। अलग-अलग वर्गों से चयन में गुणा करते हैं। / The selection is \(\binom{4}{1}\binom{3}{1}=12\). Multiply choices when selecting from different groups.
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From (5) boys and (4) girls, how many ways are there to select (2) boys?
A. (10)
केवल लड़कों में से चयन है इसलिए \(\binom{5}{2}=10\) होगा। अनावश्यक सूचना से भ्रमित न हों। / Only boys are being selected, so \(\binom{5}{2}=10\). Do not get confused by extra information.
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From (4) apples and (5) mangoes, in how many ways can (1) apple and (2) mangoes be selected?
C. (40)
तरीके \(\binom{4}{1}\binom{5}{2}=4\cdot10=40\) हैं। अलग फल वर्गों के चयन को गुणा करें। / The ways are \(\binom{4}{1}\binom{5}{2}=4\cdot10=40\). Multiply selections from separate fruit groups.
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How many lines can be formed by selecting (2) points from (8) points?
B. (28)
दो बिंदु एक रेखा बनाते हैं इसलिए \(\binom{8}{2}=28\) है। बिंदुओं का क्रम नहीं गिना जाता। / Two points form one line, so \(\binom{8}{2}=28\). The order of points is not counted.
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There are (7) points, no (3) of which are collinear. How many triangles can be formed?
B. (35)
त्रिभुज के लिए (3) बिंदु चाहिए इसलिए \(\binom{7}{3}=35\) है। सीध में न होने की शर्त त्रिभुज को संभव बनाती है। / A triangle needs (3) points, so \(\binom{7}{3}=35\). The non-collinear condition makes triangles possible.
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How many circles can be formed from (6) points if no (3) points are collinear?
C. (20)
एक वृत्त को (3) बिंदु निर्धारित करते हैं। इसलिए \(\binom{6}{3}=20\) सही है। / A circle is determined by (3) points. Hence \(\binom{6}{3}=20\) is correct.
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What is the total number of handshakes among (10) persons taken (2) at a time?
B. (45)
हर हाथ मिलाना (2) व्यक्तियों की एक जोड़ी है। इसलिए कुल संख्या \(\binom{10}{2}=45\) है। / Each handshake is a pair of (2) persons. Therefore the total number is \(\binom{10}{2}=45\).
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In how many ways can a committee of (3) members be selected from (9) members?
C. (84)
समिति चयन का प्रश्न है इसलिए \(\binom{9}{3}=84\) है। समिति में पद न हों तो क्रम नहीं गिना जाता। / This is a committee selection question, so \(\binom{9}{3}=84\). If posts are not assigned, order is not counted.
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How many ways are there to select (2) candidates for an award from (12) candidates?
B. (66)
दो समान चयनित उम्मीदवारों का क्रम महत्वपूर्ण नहीं है। अतः \(\binom{12}{2}=66\) होगा। / The order of the two selected candidates is not important. Thus \(\binom{12}{2}=66\).
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How many ways are there to choose (3) cards from (11) different cards?
C. (165)
यह \(\binom{11}{3}=165\) का सीधा प्रयोग है। कार्ड चुनने में क्रम नहीं देखा जाता। / This is a direct use of \(\binom{11}{3}=165\). Order is not considered while choosing cards.
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From (6) red balls and (4) blue balls, in how many ways can (1) red and (1) blue ball be selected?
C. (24)
तरीके \(\binom{6}{1}\binom{4}{1}=24\) हैं। अलग रंगों के चयन में गुणा नियम लगाएं। / The ways are \(\binom{6}{1}\binom{4}{1}=24\). Apply multiplication rule for selections from different colors.
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From (5) red balls and (3) black balls, how many ways are there to choose (2) red balls?
B. (10)
केवल लाल गेंदों में से चयन है। इसलिए \(\binom{5}{2}=10\) सही है। / Only red balls are being selected. Hence \(\binom{5}{2}=10\) is correct.
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How many ways are there to select (3) chairs from (7) chairs?
B. (35)
चयन में व्यवस्था नहीं करनी है इसलिए \(\binom{7}{3}=35\) है। कुर्सियां चुनना संयोजन है। / No arrangement is required, so \(\binom{7}{3}=35\). Choosing chairs is a combination.
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B. (78)
समिति के लिए केवल चयन है इसलिए \(\binom{13}{2}=78\) होगा। पद अलग न हों तो क्रम नहीं गिना जाता। / It is only selection for a committee, so \(\binom{13}{2}=78\). If posts are not different, order is not counted.
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How many ways are there to choose (4) subjects from (6) subjects?
C. (15)
\(\binom{6}{4}=\binom{6}{2}=15\) होता है। \(\binom{n}{r}=\binom{n}{n-r}\) याद रखें। / \(\binom{6}{4}=\binom{6}{2}=15\). Remember \(\binom{n}{r}=\binom{n}{n-r}\).
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What is the value of \(\binom{8}{2}\)?
C. (28)
\(\binom{8}{2}=\frac{8\cdot7}{2}=28\) है। दो वस्तुओं के चयन का सूत्र तेज होता है। / \(\binom{8}{2}=\frac{8\cdot7}{2}=28\). The formula for choosing two objects is quick.
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Find the value of \(\binom{10}{3}\).
C. (120)
\(\binom{10}{3}=\frac{10\cdot9\cdot8}{3\cdot2\cdot1}=120\) है। छोटे (r) के लिए सीधे गुणा करें। / \(\binom{10}{3}=\frac{10\cdot9\cdot8}{3\cdot2\cdot1}=120\). For small (r), multiply directly.
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What will be the value of \(\binom{9}{0}\)?
B. (1)
किसी भी (n) के लिए \(\binom{n}{0}=1\) होता है। यह संयोजन की मूल पहचान है। / For any (n), \(\binom{n}{0}=1\). This is a basic identity of combinations.
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What is the value of \(\binom{7}{7}\)?
B. (1)
सभी (7) वस्तुओं को चुनने का केवल एक तरीका है। अतः \(\binom{7}{7}=1\) है। / There is only one way to choose all (7) objects. Therefore \(\binom{7}{7}=1\).
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What is the value of \(\binom{12}{1}\)?
C. (12)
\(\binom{n}{1}=n\) होता है। इसलिए \(\binom{12}{1}=12\) सही है। / \(\binom{n}{1}=n\). Hence \(\binom{12}{1}=12\) is correct.
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Find the value of \(\binom{9}{2}\).
B. (36)
\(\binom{9}{2}=\frac{9\cdot8}{2}=36\) है। दो वस्तुओं के चयन में (2!) से भाग देना न भूलें। / \(\binom{9}{2}=\frac{9\cdot8}{2}=36\). Do not forget to divide by (2!) when choosing two objects.
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What is the value of \(\binom{11}{1}+\binom{11}{0}\)?
C. (12)
मान \(\binom{11}{1}=11\) और \(\binom{11}{0}=1\) हैं। कुल (12) होगा। / The values are \(\binom{11}{1}=11\) and \(\binom{11}{0}=1\). The total is (12).
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If \(\binom{n}{1}=14\), what is the value of (n)?
C. (14)
\(\binom{n}{1}=n\) होता है। इसलिए (n=14) मिलेगा। / \(\binom{n}{1}=n\). Therefore (n=14).
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If \(\binom{n}{0}=1\), for which (n) is it true?
C. हर \(n\geq0\)Every \(n\geq0\)
शून्य वस्तु चुनने का एक तरीका हर \(n\geq0\) के लिए होता है। यह मूल नियम है। / There is one way to choose zero objects for every \(n\geq0\). This is a basic rule.
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What is the relation between \(\binom{6}{2}\) and \(\binom{6}{4}\)?
A. दोनों बराबर हैंThey are equal
क्योंकि \(\binom{n}{r}=\binom{n}{n-r}\) होता है। यहां (r=2) और (n-r=4) हैं। / Because \(\binom{n}{r}=\binom{n}{n-r}\). Here (r=2) and (n-r=4).
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What is the value of \(\binom{5}{2}+\binom{5}{3}\)?
C. (20)
\(\binom{5}{2}=10\) और \(\binom{5}{3}=10\) हैं। इसलिए कुल (20) है। / \(\binom{5}{2}=10\) and \(\binom{5}{3}=10\). Therefore the total is (20).
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B. (12)
चयन \(\binom{3}{1}\binom{4}{1}=12\) है। अलग प्रकारों में चयन की संख्याएं गुणा होती हैं। / The selection is \(\binom{3}{1}\binom{4}{1}=12\). Choices from different types are multiplied.
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From (5) pens and (6) pencils, in how many ways can (2) pens and (1) pencil be selected?
C. (60)
तरीके \(\binom{5}{2}\binom{6}{1}=10\cdot6=60\) हैं। प्रत्येक समूह से चयन अलग-अलग करें। / The ways are \(\binom{5}{2}\binom{6}{1}=10\cdot6=60\). Select from each group separately.
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From (4) white and (5) black balls, in how many ways can (2) black balls be selected?
A. (10)
केवल काली गेंदों में से चयन है इसलिए \(\binom{5}{2}=10\) होगा। बाकी सूचना केवल संदर्भ है। / Only black balls are selected, so \(\binom{5}{2}=10\). The remaining information is only context.
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B. (28)
यह \(\binom{8}{2}=28\) है। कार्य समान हो तो चयन में क्रम नहीं गिना जाता। / This is \(\binom{8}{2}=28\). If the work is same, order is not counted in selection.
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How many ways are there to select (5) students from (6) students for a competition?
B. (6)
\(\binom{6}{5}=\binom{6}{1}=6\) है। कभी-कभी न चुने गए व्यक्ति को गिनना आसान होता है। / \(\binom{6}{5}=\binom{6}{1}=6\). Sometimes counting the person not selected is easier.
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In how many ways can a set of (3) questions be made from (10) different questions?
C. (120)
प्रश्नों का सेट बनाना चयन है। इसलिए \(\binom{10}{3}=120\) सही है। / Making a set of questions is selection. Hence \(\binom{10}{3}=120\) is correct.
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A plate is to be made by choosing (2) dishes from (5) dishes. How many plates can be made?
A. (10)
दो व्यंजन चुनने के तरीके \(\binom{5}{2}=10\) हैं। थाली में क्रम का महत्व नहीं है। / The ways to choose two dishes are \(\binom{5}{2}=10\). The order on the plate is not important.
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In how many ways can an invigilation team of (2) teachers be selected from (7) teachers?
B. (21)
टीम के लिए \(\binom{7}{2}=21\) चयन होंगे। टीम में क्रम नहीं होता। / For the team, there are \(\binom{7}{2}=21\) selections. A team has no order.
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How many ways are there to choose (4) songs from (9) songs?
C. (126)
गीत चुनना व्यवस्था नहीं है इसलिए \(\binom{9}{4}=126\) है। क्रम चाहिए होता तो अलग प्रश्न होता। / Choosing songs is not arrangement, so \(\binom{9}{4}=126\). If order were needed, it would be a different question.
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How many ways are there to choose (3) pictures from (8) different pictures for an exhibition?
B. (56)
चयन \(\binom{8}{3}=56\) होगा। प्रदर्शनी के लिए चुनना संयोजन है। / The selection is \(\binom{8}{3}=56\). Choosing for an exhibition is a combination.
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How many ways are there to choose (2) digits from (6) digits if order is not considered?
B. (15)
क्रम नहीं देखना है इसलिए \(\binom{6}{2}=15\) है। यही संयोजन और क्रमचय का मुख्य अंतर है। / Order is not considered, so \(\binom{6}{2}=15\). This is the main difference between combination and permutation.
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How many line segments can be formed from (10) points?
B. (45)
एक रेखाखंड के लिए (2) बिंदु चाहिए। इसलिए \(\binom{10}{2}=45\) है। / A line segment needs (2) points. Therefore \(\binom{10}{2}=45\).
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In how many ways can (3) letters be selected from (8) letters if order is not important?
B. (56)
क्रम महत्वपूर्ण नहीं है इसलिए \(\binom{8}{3}=56\) है। चयन वाले प्रश्नों में संयोजन लगाएं। / Order is not important, so \(\binom{8}{3}=56\). Use combinations in selection questions.
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From (5) boys and (5) girls, in how many ways can (1) boy and (1) girl be selected?
C. (25)
तरीके \(\binom{5}{1}\binom{5}{1}=25\) हैं। दोनों चयन स्वतंत्र हैं इसलिए गुणा करें। / The ways are \(\binom{5}{1}\binom{5}{1}=25\). The two selections are independent, so multiply.
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How many ways are there to choose (4) toys from (12) toys?
C. (495)
यह \(\binom{12}{4}=495\) है। बड़े मानों में सूत्र को क्रम से सरल करें। / This is \(\binom{12}{4}=495\). For larger values, simplify the formula step by step.
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In how many ways can at least (1) pencil be selected from (7) colored pencils?
C. (127)
कुल उपसमुच्चय \(2^7\) हैं और खाली चयन हटेगा। इसलिए \(2^7-1=127\) तरीके हैं। / Total subsets are \(2^7\), and the empty selection is removed. Hence the ways are \(2^7-1=127\).
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