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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exemplar Solution

A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?

[Hint: Required number of ways = ³C₁ × ⁶C₂ + ³C₂ × ⁶C₂+ ³C₃.]

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e 2 w h i t e, 3 b l a c k a n d 4 r e d b a l l s i n a b o x . 3 b a l l s a r e t o b e d r a w n o u t o f 9 b a l l s \\ a t l e a s t o n e b l a c k b a l l i s t o b e i n c l u d e d . \\ S o, t h e p o s s i b l e s e l e c t i o n i s \\ (1 b l a c k a n d 2 o t h e r b a l l s) o r (2 b l a c k a n d 1 o t h e r b a l l) o r (3 b l a c k a n d n o o t h e r b a l l) \\ S o, t h e n u m b e r o f p o s s i b l e s e l e c t i o n i s \\ = {}_{3}{C_{1}} \times {}_{6}{C_{2}} + {}_{3}{C_{2}} \times {}_{6}{C_{1}} + {}_{3}{C_{3}} \times {}_{6}{C_{0}} \\ = 3 \times 15 + 3 \times 6 + 1 \times 1 = 45 + 18 + 1 = 64 \\ H e n c e, t h e r e q u i r e d s e l e c t i o n = 64 . \end{array}$

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9 months ago

Banasthali Vidyapith JEE

Can I get the IT branch at Banasthali Vidyapith with 1180 rank in JEE?

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

Find the coefficient of $x^{15}$ in the expansion of ${(x - x^{2})}^{10} .$

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} T h e g i v e n expression i s {(x - x^{2})}^{10} \\ G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ = {}_{10}{C_{r}} {(x)}^{10 - r} {(- x^{2})}^{r} = {}_{10}{C_{r}} {(x)}^{10 - r} {(- 1)}^{r} {(x^{2})}^{r} \\ = {(- 1)}^{r} . {}_{10}{C_{r}} {(x)}^{10 - r + 2 r} = {(- 1)}^{r} . {}_{10}{C_{r}} {(x)}^{10 + r} \\ T o f i n d t h e c o e f f i c i e n t o f x^{15}, P u t 10 + r = 15 \Rightarrow r = 5 \\ ∴ C o e f f i c i e n t o f x^{15} = {(- 1)}^{5} . {}_{10}{C_{5}} = - {}_{10}{C_{5}} = - 252 \\ H e n c e, t h e r e q u i r e d c o e f f i c i e n t = - 252 \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exemplar Solution

There are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated.
[Hint: Required number = 2¹⁰ – 1.]

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f l a m p s = 10 \\ T h e t o t a l n u m b e r o f w a y s i n w h i c h h a l l c a n b e illuminated i s e q u a l t o t h e n u m b e r o f \\ s e l e c t i o n o f o n e o r m o r e i t e m s o u t o f n d i f f e r e n t i t e m s . \\ i . e ., {}_{n}{C_{1}} + {}_{n}{C_{2}} + {}_{n}{C_{3}} + {}_{n}{C_{4}} + \dots + {}_{n}{C_{n}} = 2^{n} - 1 \\ F r o m B i n o m i a l expansion, w e h a v e \\ {}_{n}{C_{0}} + {}_{n}{C_{1}} + {}_{n}{C_{2}} + {}_{n}{C_{3}} + {}_{n}{C_{4}} + \dots + {}_{n}{C_{n}} = 2^{n} \\ S o t o t a l n u m b e r o f w a y s = {}_{10}{C_{1}} + {}_{10}{C_{2}} + {}_{10}{C_{3}} + \dots + {}_{10}{C_{10}} \\ = 2^{10} - 1 = 1024 - 1 = 1023 \\ H e n c e, t h e r e q u i r e d n u m b e r o f p o s s i b l e w a y s = 1023 . \end{array}$

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9 months ago

Indian Institute for Aeronautical Engineering & Information Technology (IIAEIT) Colleges in Pune

What is the fees of studying BTech at IIAEIT?

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Sejal Baveja

Contributor-Level 10

The total tuition fees of BTech at Indian Institute for Aeronautical Engineering and Information Technology is INR 9.2 lakh. The total fees include other components like hostel fees, security deposits, etc. Candidates are required to pay the fees or a certain amount of it to secure their seat in the course.

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9 months ago

Architecture Entrance Exam Exam

Is it necessary to send the hard copy of the application form to the NATA office?

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

Find the middle term(s) in the expansion of:

${(x / a - \frac{a}{x})}^{10} .$ 3x – x³/6 ⁹

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} (i) G i v e n expression i s {(\frac{x}{a} - \frac{a}{x})}^{10} \\ N u m b e r o f t e r m s = 10 + 1 = 11 (o d d) \\ ∴ M i d d l e t e r m {(\frac{n + 1}{2})}^{t h} t e r m = \frac{11 + 1}{2} = \frac{12}{2} = 6^{t h} t e r m \\ G e n e r a l t e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ \Rightarrow T_{5 + 1} = {}_{10}{C_{5}} {(\frac{x}{a})}^{10 - 5} {(- \frac{a}{x})}^{5} = - {}_{10}{C_{5}} \frac{x^{5}}{a^{5}} . \frac{a^{5}}{x^{5}} = - {}_{10}{C_{5}} \\ = - \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} . = - 9 \times 7 \times 4 = - 252 \\ H e n c e, t h e r e q u i r e d m i d d l e t e r m = - 252 \\ (i i) G i v e n expression i s {(3 x - \frac{x^{3}}{6})}^{9} \\ N u m b e r o f t e r m s = 9 + 1 = 10 (e v e n) \\ ∴ M i d d l e t e r m s a r e {(\frac{n}{2})}^{t h} t e r m a n d {(\frac{n}{2} + 1)}^{t h} t e r m \\ {(\frac{10}{2})}^{t h} = 5^{t h} t e r m a n d {(\frac{10}{2} + 1)}^{t h} = 6^{t h} t e r m \\ G e n e r a l t e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ ∴ T_{5} = T_{4 + 1} = {}_{9}{C_{4}} {(3 x)}^{9 - 4} {(- \frac{x^{3}}{6})}^{4} = {}_{9}{C_{4}} {(3)}^{5} . x^{5} {(- \frac{1}{6})}^{4} . x^{12} \\ = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times \frac{3 \times 3 \times 3 \times 3 \times 3}{6 \times 6 \times 6 \times 6} x^{17} = \frac{189}{8} x^{17} \\ N o w, T_{6} = T_{5 + 1} = {}_{9}{C_{5}} {(3 x)}^{9 - 5} {(- \frac{x^{3}}{6})}^{5} \\ = {}_{9}{C_{5}} {(3)}^{4} . x^{4} {(- \frac{1}{6})}^{5} . x^{15} \\ = \frac{9 \times 8 \times 7 \times 6 \times 5}{5 \times 4 \times 3 \times 2 \times 1} {(3)}^{4} {(- \frac{1}{6})}^{5} . x^{19} = - \frac{21}{16} x^{19} \\ H e n c e, t h e r e q u i r e d m i d d l e t e r m s a r e \frac{189}{8} x^{17} a n d - \frac{21}{16} x^{19} \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exemplar Solution

There are 10 persons named P₁, P₂, P₃, ... P₁₀. Out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement P₁ must occur whereas P₄ and P₅ do not occur. Find the number of such possible arrangements.

[Hint: Required number of arrangement = ⁷C₄ × 5!]

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t P_{1}, P_{2}, P_{3}, P_{4}, \dots, P_{10} a r e 10 p e r s o n s o u t o f w h i c h 5 p e r s o n s a r e t o b e a r r a n g e d b u t \\ P_{1} m u s t o c c u r a n d P_{4} a n d P_{5} n e v e r o c c u r . \\ ∴ s e l e c t i o n i s t o b e d o n e o n l y f o r 10 - 3 = 7 p e r s o n s \\ ∴ N u m b e r o f s e l e c t i o n = {}_{7}{C_{4}} = \frac{7!}{4! (7 - 4)!} = \frac{7!}{4! 3!} = \frac{7.6.5.4!}{4! . 3.2.1} = 35 \\ 5 p e o p l e c a n b e a r r a n g e d a s 5! \\ S o, t h e n u m b e r o f a r r a n g e m e n t = 35 \times 5! = 35 \times 120 = 4200 \\ H e n c e, t h e r e q u i r e d a r r a n g e m e n t = 4200 . \end{array}$

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9 months ago

Which university is more affordable for BTech course among Kerala Agricultural University and TNAU?

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Tanisha Pandey

Contributor-Level 10

The total tuition fee for a BTech course at Kerala Agricultural University ranges from INR 60,000 to INR 2.1 lakh. Moreover, the tuition fee for BTech at TNAU is INR 36,000. Therefore, it can be explicitly seen that the tuition fee for the BTech course at TNAU is more affordable compared to Kerala Agricultural University. The fee amount mentioned is sourced from the official website/sanctioning body and is subject to change. Therefore, it is indicative.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

Find the term independent of $x$ in the expansion of 3x – 2/x^{2 15}

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} G i v e n expression i s {(3 x - \frac{2}{x^{2}})}^{15} \\ G e n e r a l t e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ = {}_{15}{C_{r}} {(3 x)}^{15 - r} {(- \frac{2}{x^{2}})}^{r} = {}_{15}{C_{r}} {(3)}^{15 - r} {(x)}^{15 - r} {(- 2)}^{r} . \frac{1}{x^{2 r}} \\ = {}_{15}{C_{r}} {(3)}^{15 - r} . {(x)}^{15 - r - 2 r} {(- 2)}^{r} \\ = {}_{15}{C_{r}} {(3)}^{15 - r} {(x)}^{15 - 3 r} {(- 2)}^{r} \\ f o r g e t t i n g t h e t e r m i n d e p e n d e n t o f x, \\ 15 - 3 r = 0 \Rightarrow r = 5 \\ O n p u t t i n g t h e v a l u e o f r i n t h e a b o v e expression, w e g e t \\ = {}_{15}{C_{5}} {(3)}^{15 - 5} {(- 2)}^{5} = - {}_{15}{C_{5}} {(3)}^{10} {(2)}^{5} \\ = - \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} . {(3)}^{10} {(2)}^{5} = - 7 \times 13 \times 3 \times 11 . {(3)}^{10} {(2)}^{5} = - 3003 . {(3)}^{10} {(2)}^{5} \\ H e n c e, t h e r e q u i r e d t e r m = - 3003 . {(3)}^{10} {(2)}^{5} \end{array}$

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9 months ago

Elphinstone College B.Sc.

Is there any entrance exam for BSc admissions at Elphinstone College?

0 Follower 7 Views

Saurabh Khanduri

Contributor-Level 10

For admissions into Elphinstone College, candidates are not required to qualify for any exam. All a candidate needs to do to take admission in BSc at Elphinstone College is to apply for admission on the official portal of Homi Bhabha State University.

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9 months ago

IIT Kharagpur - Indian Institute of Technology B.Tech

Can I get admission to IIT Kharagpur with a rank of 10,000?

0 Follower 21 Views

Neha Arora

Contributor-Level 10

IIT Kharagpur JEE Advanced cutoff 2025 has been released for round 3. The cutoff was released category-wise for students seeking admission to the BTech/Integrated BTech/BSc specialisations at the college. For candidates belonging to the General AI category, the cutoff ranks varied between 450 and 18182. As per the cutoff range, it is possible to get admission to the college at 10,000 rank in JEE Advanced. To check which specialisation is available at this rank, candidates can refer to the table below. The table below consist of round 3 cutoff ranks for the General AI category candidates in some subjects:

Course	2025
B.Tech. (Hons.) in Aerospace Engineering	4567
B.Tech. (Hons.) in Agricultural and Food Engineering	10498
B.Tech. (Hons.) in Biotechnology and Biochemical Engineering	8070
B.Tech. (Hons.) in Civil Engineering	7180
B.Tech. (Hons.) in Chemical Engineering	4820
B.Tech. (Hons.) in Computer Science and Engineering	450
B.Tech. (Hons.) in Electronics and Electrical Communication Engineering	1454
B.Tech. (Hons.) in Electrical Engineering	1898
B.Tech. (Hons.) in Instrumentation Engineering	2561
B.Tech. (Hons.) in Industrial Engineering	4113
B.Tech. (Hons.) in Mechanical Engineering	3672
B.Tech. (Hons.) in Manufacturing Science and Engineering	6117
B.Tech. (Hons.) in Metallurgical and Materials Engineering	7078
B.Tech. (Hons.) in Ocean Engineering and Naval Architecture	9020
B.Tech. (Hons.) in Mining Engineering	9726
B.Sc. in Chemistry	11933
B.Sc. in Economics	5845
B.Sc. in Physics	9103
B.Sc. in Mathematics and Computing	1335

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exemplar Solution

Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided that no digit is to be repeated.

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} A n y n u m b e r d i v i s i b l e b y 5, i t s u n i t p l a c e m u s t h a v e 0 o r 5 \\ W e h a v e t o f i n d 4 - d i g i t n u m b e r g r e a t e r t h a n 6000 a n d l e s s t h a n 7000 . \\ S o, t h e u n i t p l a c e c a n b e f i l l e d w i t h 2 w a y s (0 o r 5) since, r e p e t i t i o n i s n o t a l l o w e d . \\ ∴ t e n s p l a c e c a n b e f i l l e d w i t h 7 w a y s a n d h u n d r e d s p l a c e c a n b e f i l l e d w i t h 8 w a y s . \\ B u t t h e r e q u i r e d n u m b e r i s g r e a t e r t h a n 6000 a n d l e s s t h a n 7000 . \\ S o, t h o u s a n d p l a c e c a n b e f i l l e d w i t h 1 d i g i t s i . e . 6 \\ \underset{1}{T h} \underset{8}{H} \underset{7}{T} \underset{2}{O} \\ S o, t h e t o t a l n u m b e r o f integers = 1 \times 8 \times 7 \times 2 = 112 \\ H e n c e, t h e r e q u i r e d n u m b e r o f integers = 112 \end{array}$

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9 months ago

Architecture Entrance Exam Exam

What documents are required while filling the NATA application form?

0 Follower 8 Views

Parul Pandey

Contributor-Level 7

Recent passport-size photograph
Signature (scanned)
Valid ID proof (Aadhar/Passport/etc.)
Educational certificates
Caste/Disability certificate (if applicable)

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

Find the coefficient of $x$ in the expansion of

$(1 - 3 x + 7 x^{2}) {(1 - x)}^{16} .$

0 Follower 17 Views

Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} T h e g i v e n expression i s (1 - 3 x + 7 x^{2}) {(1 - x)}^{16} \\ = (1 - 3 x + 7 x^{2}) [{}_{16}{C_{0}} {(1)}^{16} {(- x)}^{0} + {}_{16}{C_{1}} {(1)}^{15} (- x) + {}_{16}{C_{2}} {(1)}^{14} {(- x)}^{2} + \dots] \\ = (1 - 3 x + 7 x^{2}) (1 - 16 x + 120 x^{2} + \dots) \\ C o l l e c t i n g t h e t e r m c o n t a i n i n g x, w e g e t - 16 x - 3 x = - 19 x \\ H e n c e, t h e c o e f f i c i e n t o f x = - 19 \end{array}$

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9 months ago

Elphinstone College B.Sc.

Can I enrol in the BSc courses at Elphinstone College after clearing Class 10 Board Exams?

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Saurabh Khanduri

Contributor-Level 10

No, to be eligible for admissions in BSc courses at Elphinstone College, candidates must clear their Class 12 board exams. Candidates who have cleared their Class 12 board exams are eligible to apply for BSc courses at Elphinstone College if they have cleared their examination with an aggregate of 45% from any stream.

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9 months ago

Teaching & Education UGC NET Notifications

Is the UGC NET 2026 June session notification released?

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Chanchal Shekhar

Contributor-Level 6

The NTA is expected to release the UGC NET notification in April 2026. It will be announced in the form of PDF at ugcnet.nta.nic.in. Along with it, application link will also be activated.

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9 months ago

Drexel University Study Abroad

What are the overall rankings at Drexel University?

0 Follower 5 Views

Tasneem Hoda

Contributor-Level 10

According to the Shiksha Popularity Rankings, Drexel University has been ranked at #93 at the country level and #411 at the global level. Some of the worldwide rankings of the university are given in the table below:

Ranking Body Name	2024	2025	2026
Shiksha Popularity Rankings	364	411	NA
Times Higher Education	301	351	NA
US News and World Report	428	493	NA
Quacquarelli Symonds	601	721	711

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

If the term free from $x$ in the expansion of x – k/x² ¹⁰ is 405, find the value of $k$ .

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} T h e g i v e n expression i s {(\sqrt[]{x} - \frac{K}{x^{2}})}^{10} \\ G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ = {}_{10}{C_{r}} {(\sqrt[]{x})}^{10 - r} {(\frac{- K}{x^{2}})}^{r} = {}_{10}{C_{r}} {(x)}^{\frac{10 - r}{2}} {(- K)}^{r} (\frac{1}{x^{2 r}}) \\ = {}_{10}{C_{r}} {(x)}^{\frac{10 - r}{2} - 2 r} {(- K)}^{r} = {}_{10}{C_{r}} {(x)}^{\frac{10 - r - 4 r}{2}} {(- K)}^{r} \\ = {}_{10}{C_{r}} {(x)}^{\frac{10 - 5 r}{2}} {(- K)}^{r} \\ f o r g e t t i n g t h e t e r m i n d e p e n d e n t o f x, \\ \frac{10 - 5 r}{2} = 0 \Rightarrow r = 2 \\ O n p u t t i n g t h e v a l u e o f r i n t h e a b o v e expression, w e g e t \\ = {}_{10}{C_{2}} {(- K)}^{2} \\ A c c o r d i n g t o t h e c o n d i t i o n o f t h e q u e s t i o n, w e h a v e \\ {}_{10}{C_{2}} K^{2} = 405 \Rightarrow \frac{10.9}{2.1} K^{2} = 405 \\ \Rightarrow 45 K^{2} = 405 \Rightarrow K^{2} = \frac{405}{45} = 9 \\ ∴ K = \pm 3 \\ H e n c e, t h e v a l u e o f K = \pm 3 \end{array}$

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9 months ago

Elphinstone College B.Sc.

How many specialisations of BSc courses are available at Elphinstone College?

0 Follower 3 Views

Saurabh Khanduri

Contributor-Level 10

The BSc course at Elphinstone College is available in 12 different specialisations. These include Zoology, Information Technology, Chemistry, Microbiology, Biotechnology, and several others. Out of all of these specialisations, the BSc course in Chemistry is highly popular among applicants.

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