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Exemplar Solution Class 11th

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.]

5 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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

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

6 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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!]

7 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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

6 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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.

8 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

Find the coefficient of $x$ in the expansion of

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

17 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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

4 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

Find the number of different words that can be formed from the letters of the word ‘TRIANGLE’ so that no vowels are together.

11 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} T o t a l n u m b e r o f w o r d s i n' T R I A N G L E' = 8 \\ O u t o f 5 a r e c o n s o n a n t s a n d 3 a r e v o w e l s \\ I f v o w e l s a r e n o t t o g e t h e r, t a k e n w e h a v e t h e f o l l o w i n g a r r a n g e m e n t \\ V Unsupported ? C Unsupported ? V Unsupported ? C Unsupported ? V Unsupported ? C Unsupported ? V Unsupported ? C Unsupported ? V Unsupported ? C Unsupported ? V \\ c o n s o n a n t c a n b e a r r a n g e d i n 5! = 120 w a y s \\ V o w e l o c c u p y 6 p l a c e s \\ ∴ 3 v o w e l s c a n b e a r r a n g e d i n 6 p l a c e s = {}_{6}{P_{3}} \\ = \frac{6!}{(6 - 3)!} = \frac{6!}{3!} = 120 w a y s \\ S o, t h e t o t a l a r r a n g e m e n t = 120 \times 120 = 14400 w a y s \\ H e r e, t h e r e q u i r e d a r r a n g e m e n t = 14400 w a y s . \end{array}$

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Exemplar Solution Class 11th

Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.

3 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} T o t a l n u m b e r o f t h i n g s = n \\ 3 t h i n g s m u s t b e t o g e t h e r \\ ∴ t h e n u m b e r o f r e m a i n i n g t h i n g s = n - 3 \\ N u m b e r o f t h i n g s t o b e s e l e c t e d = r \\ O u t o f r, 3 a r e a l w a y s t o g e t h e r \\ ∴ N u m b e r o f w a y s o f s e l e c t i o n = {}_{n - 3}{C_{r - 2}} \\ N o w p e r m u t a t i o n o f 3 t h i n g s w h i c h a r e a l w a y s t o g e t h e r = 3! \\ N u m b e r o f p e r m u t a t i o n o f (r - 2) t h i n g s = (r - 2)! \\ ∴ T o t a l n u m b e r o f a r r a n g e m e n t s = {}_{n - 3}{C_{r - 2}} \times (r - 2)! \times 3! \\ H e n c e, t h e r e q u i r e d a r r a n g e m e n t s = {}_{n - 3}{C_{r - 2}} \times (r - 2)! \times 3! . \end{array}$

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Exemplar Solution Class 11th

Find the term independent of $x$ , $x \neq 0$ , in the expansion of (3x²/2 – 1/3x)¹⁵

6 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ = {}_{15}{C_{r}} {(\frac{3 x^{2}}{2})}^{15 - r} {(- \frac{1}{3 x})}^{r} = {}_{15}{C_{r}} {(\frac{3}{2})}^{15 - r} {(x)}^{30 - 2 r} {(- \frac{1}{3})}^{r} \frac{1}{x^{r}} \\ = {}_{15}{C_{r}} {(\frac{3}{2})}^{15 - r} {(x)}^{30 - 2 r - r} {(- 1)}^{r} \frac{1}{{(3)}^{r}} \\ = {}_{15}{C_{r}} {(\frac{3}{2})}^{15 - r} {(x)}^{30 - 3 r} {(- 1)}^{r} \frac{1}{{(3)}^{r}} \\ f o r g e t t i n g t h e t e r m i n d e p e n d e n t o f x, \\ 30 - 3 r = 0 \Rightarrow r = 10 \\ O n p u t t i n g t h e v a l u e o f r i n t h e a b o v e expression, w e g e t \\ = {}_{15}{C_{10}} {(\frac{3}{2})}^{15 - 10} {(- 1)}^{10} \frac{1}{{(3)}^{10}} = {}_{15}{C_{10}} \frac{{(3)}^{5}}{{(2)}^{5}} . \frac{1}{{(3)}^{10}} \\ = {}_{15}{C_{10}} . \frac{1}{{(2)}^{5} . {(3)}^{5}} = {}_{15}{C_{10}} {(\frac{1}{6})}^{5} \\ H e n c e, t h e r e q u i r e d t e r m = {}_{15}{C_{10}} {(\frac{1}{6})}^{5} \end{array}$

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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.]

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

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

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!]

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

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.

Find the coefficient of $x$ in the expansion of

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

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

Find the number of different words that can be formed from the letters of the word ‘TRIANGLE’ so that no vowels are together.

Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.

Find the term independent of $x$ , $x \neq 0$ , in the expansion of (3x²/2 – 1/3x)¹⁵