Class 11th

Find the sixth term of the expansion ${(y^{\frac{1}{2}} + x^{\frac{1}{3}})}^{n}$ , if the binomial coefficient of the third term from the end is 45.

[Hint: Binomial coefficient of third term from the end = Binomial coefficient of third term from the beginning = ⁿC₂.]

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Contributor-Level 10

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

$\begin{array}{l} T h e g i v e n expression i s {(y^{\frac{1}{2}} + x^{\frac{1}{3}})}^{n}, since t h e b i n o m i a l c o e f f i c i e n t o f t h i r d \\ t e r m f r o m t h e e n d = b i n o m i a l c o e f f i c i e n t o f t h i r d t e r m f r o m t h e b e g i n n i n g = {}_{n}{C_{2}} \\ ∴ {}_{n}{C_{2}} = 45 \\ \Rightarrow \frac{n (n - 1)}{2} = 45 \Rightarrow n^{2} - n = 90 \\ \Rightarrow n^{2} - n - 90 = 0 \Rightarrow n^{2} - 10 n + 9 n - 90 = 0 \\ \Rightarrow n (n - 10) + 9 (n - 10) = 0 \Rightarrow (n - 10) (n + 9) = 0 \\ \Rightarrow n = 10, n = - 9 \Rightarrow n = 10, n \neq - 9 \\ S o, t h e g i v e n expression b e c o m e s {(y^{\frac{1}{2}} + x^{\frac{1}{3}})}^{10} \\ S i x t h t e r m i n t h i s expression \\ T_{6} = T_{5 + 1} = {}_{10}{C_{5}} {(y^{\frac{1}{2}})}^{10 - 5} {(x^{\frac{1}{3}})}^{5} = {}_{10}{C_{5}} y^{\frac{5}{2}} . x^{\frac{5}{3}} \\ = 252 y^{\frac{5}{2}} . x^{\frac{5}{3}} \\ H e n c e, t h e r e q u i r e d t e r m = 252 y^{\frac{5}{2}} . x^{\frac{5}{3}} \end{array}$

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New answer posted

a year ago

If ⁿC_{r – 1} = 36, ⁿC_r = 84, and ⁿC_{r + 1}= 126, then find ^rC₂.

[Hint: Form equation using C(n, r-1), C(n, r), and C(n, r+1) to find the value of r.]

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Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Nine

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t {}_{n}{C_{r - 1}} = 36 \dots (i) \\ {}_{n}{C_{r}} = 84 \dots (i i) \\ {}_{n}{C_{r + 1}} = 126 \dots (i i i) \\ D i v i d i n g e q n . (i) b y e q n . (i i) w e g e t \\ \frac{{}_{n}{C_{r - 1}}}{{}_{n}{C_{r}}} = \frac{36}{84} \\ \Rightarrow \frac{\frac{n!}{(r - 1)! (n - r + 1)!}}{\frac{n!}{r! (n - r)!}} = \frac{3}{7} [? {}_{n}{C_{r}} = \frac{n!}{r! (n - r)!}] \\ \Rightarrow \frac{n!}{(r - 1)! (n - r + 1)!} * \frac{r! (n - r)!}{n!} = \frac{3}{7} \\ \Rightarrow \frac{r . (r - 1)! (n - r)!}{(r - 1)! (n - r + 1)! + (n - r)!} = \frac{3}{7} \\ \Rightarrow \frac{r}{n - r + 1} = \frac{3}{7} \Rightarrow 3 n - 3 r + 3 = 7 r \\ \Rightarrow 3 n - 10 r = - 3 \dots (i v) \\ D i v i d i n g e q n . (i i) b y e q n . (i i i) w e g e t \\ \frac{{}_{n}{C_{r}}}{{}_{n}{C_{r + 1}}} = \frac{84}{126} \Rightarrow \frac{\frac{n!}{r! (n - r)!}}{\frac{n!}{(r + 1)! (n - r - 1)!}} = \frac{2}{3} \\ \Rightarrow &th \end{array}$

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New answer posted

a year ago

The first term of an A.P. is $a$ , and the sum of the first $p$ terms is zero, show that the

sum of its next $q$ terms is $\frac{- a (p + q) q}{p - 1}$ .

[Hint: Required sum = $S_{p + q} - S_{p}$ ]

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t a_{1} = a a n d S_{p} \\ S u m o f n e x t q t e r m s o f t h e g i v e n A . P . = S_{p + q} - S_{p} \\ ∴ S_{p + q} = \frac{p + q}{2} [2 a + (p + q - 1) d] \\ a n d S_{p} = \frac{p}{2} [2 a + (p - 1) d] = 0 \\ \Rightarrow 2 a + (p - 1) d = 0 \Rightarrow (p - 1) d = - 2 a \\ \Rightarrow d = \frac{- 2 a}{p - 1} \\ S u m o f n e x t q t e r m s = S_{p + q} - S_{p} \\ = \frac{p + q}{2} [2 a + (p + q - 1) d] - 0 \\ = \frac{p + q}{2} [2 a + (p + q - 1) (\frac{- 2 a}{p - 1})] \\ = \frac{p + q}{2} [2 a + \frac{(p - 1) (- 2 a)}{p - 1} - \frac{2 a q}{p - 1}] \\ = \frac{p + q}{2} [2 a - 2 a - \frac{2 a q}{p - 1}] = \frac{p + q}{2} (- \frac{2 a q}{p - 1}) \\ = \frac{- a (p + q) q}{p - 1} \\ H e n c e, t h e r e q u i r e d s u m = \frac{- a (p + q) q}{p - 1} \end{array}$

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New answer posted

a year ago

Find the coefficient of $\frac{1}{x^{17}}$ in the expansion of x⁴– 1/x^{3 15}

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Contributor-Level 10

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

$\begin{array}{l} T h e g i v e n expression i s {(x^{4} - \frac{1}{x^{3}})}^{15} \\ G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ = {}_{15}{C_{r}} {(x^{4})}^{15 - r} {(- \frac{1}{x^{3}})}^{r} = {}_{15}{C_{r}} {(x)}^{60 - 4 r} {(- 1)}^{r} . \frac{1}{{x^{3}}^{r}} \\ = {}_{15}{C_{r}} {(- 1)}^{r} . \frac{1}{{x^{3}}^{r - 60 + 4 r}} = {}_{15}{C_{r}} {(- 1)}^{r} . \frac{1}{{x^{7}}^{r - 60}} \\ T o f i n d t h e c o e f f i c i e n t o f \frac{1}{x^{17}}, P u t 7 r - 60 = 17 \Rightarrow 7 r = 60 + 11 \Rightarrow r = \frac{77}{7} = 11 \\ P u t t i n g t h e v a l u e o f r i n t h e a b o v e expression, w e g e t \\ = {}_{15}{C_{11}} {(- 1)}^{11} . \frac{1}{x^{17}} = - {}_{15}{C_{4}} . \frac{1}{x^{17}} = - \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1} . \frac{1}{x^{17}} \\ = - 1365 . \frac{1}{x^{17}} \\ H e n c e, t h e r e q u i r e d c o e f f i c i e n t o f \frac{1}{x^{17}} = - 1365 \end{array}$

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New answer posted

a year ago

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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Contributor-Level 10

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

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

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New answer posted

a year ago

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

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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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New answer posted

a year ago

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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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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New answer posted

a year ago

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

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

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Contributor-Level 10

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

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

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New answer posted

a year ago

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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Contributor-Level 10

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

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

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New answer posted

a year ago

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

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