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New answer posted•a year ago

Exemplar Solution Class 11th

The number of different words that can be formed from the letters of the word INTERMEDIATE such that two vowels never come together is ______.
[Hint: Number of ways of arranging 6 consonants of which two are alike is  $\frac{6!}{2!}$ and number of ways of arranging vowels $7 P 6 = \frac{3! \times 2!}{1! \times 2!}$ .]

9 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f w o r d s i s I N T E R M E D I A T E = 12 \\ w h i c h h a v e 6 v o w e l s a n d 6 c o n s o n a n t s \\ I f t w o v o w e l s n e v e r c o m e t o g e t h e r t h e n w e c a n a r r a n g e a s u n d e r \\ V C V C V C V C V C V C V \\ H e r e, v o w e l s a r e I E E I A E w h e r e 2 I' s a n d 3 E' s a r e t h e r e . \\ ∴ N u m b e r o f w a y s o f a r r a n g i n g v o w e l s = \frac{7!}{3! 2!} = 420 \\ c o n s o n a n t s a r e N T R M D T w h e r e 2 T' s a r e t h e r e . \\ ∴ N u m b e r o f w a y s o f a r r a n g i n g c o n s o n a n t s = \frac{6!}{2!} = \frac{6.5.4.3.2!}{2!} = 360 \\ S o, t h e t o t a l n u m b e r o f w o r d s a r e = 420 \times 360 = 151200 . \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 151200 . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

Let $S_{n}$ denote the sum of the first $n$ terms of an A.P. If $S_{2 n} = 3 S_{n}$ , then $S_{3 n} : S_{n}$ is equal to:

(a) 4

(b) 6

(c) 8

(d) 10

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ ∴ S_{2 n} = \frac{2 n}{2} [2 a + (2 n - 1) d] \\ S_{3 n} = \frac{3 n}{2} [2 a + (3 n - 1) d] \\ A s p e r t h e c o n d i t i o n o f t h e q u e s t i o n, w e h a v e \\ S_{2 n} = 3 . S_{n} \\ \Rightarrow \frac{2 n}{2} [2 a + (2 n - 1) d] = 3 . \frac{n}{2} [2 a + (n - 1) d] \\ \Rightarrow 2 [2 a + (2 n - 1) d] = 3 [2 a + (n - 1) d] \\ \Rightarrow 4 a + (4 n - 2) d = 6 a + (3 n - 3) d \\ \Rightarrow 6 a + (3 n - 3) d - 4 a - (4 n - 2) d = 0 \\ \Rightarrow 2 a + (3 n - 3 - 4 n + 2) d = 0 \Rightarrow 2 a + (- n - 1) d = 0 \\ \Rightarrow 2 a - (n + 1) d = 0 \Rightarrow 2 a = (n + 1) d \dots (i) \\ N o w S_{3 n} : S_{n} = \frac{3 n}{2} [2 a + (3 n - 1) d] : \frac{n}{2} [2 a + (n - 1) d] \\ = \frac{\frac{3 n}{2} [2 a + (3 n - 1) d]}{\frac{n}{2} [2 a + (n - 1) d]} = \frac{3 [2 a + (3 n - 1) d]}{2 a + (n - 1) d} \\ = \frac{3 [(n + 1) d + (3 n - 1) d]}{(n + 1) d + (n - 1) d} \\ = \frac{3 d [n + 1 + 3 n - 1]}{d (n + 1 + n - 1) d} = \frac{3 [4 n]}{2 n} = 6 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

The number of permutations of $n$ different objects, taken $r$ at a time, when repetitions are allowed, is ______.

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$N u m b e r o f p e r m u t a t i o n o f n d i f f e r e n t o b j e c t s, t a k e n r a t a t i m e i s n^{r} .$

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New question posted•a year ago

Exemplar Solution Class 11th

The number of permutations of  different objects, taken  at a time, when repetitions are allowed, is ______.

4 Views|Posted a year ago

Asked by Payal Gupta

New question posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

Let   denote the sum of the first  terms of an A.P. If , then  is equal to: (a) 4 (b) 6 (c) 8 (d) 10

3 Views|Posted a year ago

Asked by Alok Kumar Singh

New answer posted•a year ago

Exemplar Solution Class 11th

¹⁵C₈ + ¹⁵C₉ – ¹⁵C₆ – ¹⁵C₇ = ______.

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} {}_{15}{C_{8}} + {}_{15}{C_{9}} - {}_{15}{C_{6}} - {}_{15}{C_{7}} = {}_{15}{C_{15 - 8}} + {}_{15}{C_{15 - 9}} - {}_{15}{C_{6}} - {}_{15}{C_{7}} [? {}_{n}{C_{r}} = {}_{n}{C_{n - r}}] \\ = {}_{15}{C_{7}} + {}_{15}{C_{6}} - {}_{15}{C_{6}} - {}_{15}{C_{7}} = 0 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 0 . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

If in an A.P. $S_{n} = q n^{2}$ ,  and $S_{m} = q m^{2}$ , where $S_{r}$ denotes the sum of $r$ terms of the A.P., then $S_{q}$ equals

(a) $\frac{q^{3}}{2}$ 

(b) $m n q$ 

(c) $q^{3}$ 

(d) $(m + n) q^{2}$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n s e r i e s i s A . P . w h o s e f i r s t t e r m i s a a n d c o m m o n d i f f e r e n c e i s d . \\ ∴ S_{n} = \frac{n}{2} [2 a + (n - 1) d] = q n^{2} \\ \Rightarrow = 2 a + (n - 1) d = 2 q n \dots (i) \\ S_{m} = \frac{m}{2} [2 a + (m - 1) d] = q m^{2} \\ \Rightarrow = 2 a + (m - 1) d = 2 q m \dots (i i) \\ S o l v i n g e q n . (i) a n d e q n . (i i) w e g e t \\ 2 a + (m - 1) d = 2 q m \\ \underline{2 a + (n - 1) d = 2 q n} \\ (-) (-) (-) \\ (m - n) d = 2 q m - 2 q n \\ (m - n) d = 2 q (m - n) \\ ∴ d = 2 q \\ P u t t i n g t h e v a l u e o f d f r o m e q n . (i i), w e g e t \\ 2 a + (m - 1) . 2 q = 2 q m \Rightarrow 2 a = 2 q m - (m - 1) 2 q \\ \Rightarrow 2 a = 2 q (m - m + 1) \Rightarrow 2 a = 2 q \Rightarrow a = q \\ ∴ S_{q} = \frac{q}{2} [2 a + (q - 1) d] = \frac{q}{2} [2 q + (q - 1) 2 q] \\ = \frac{q}{2} [2 q + 2 q^{2} - 2 q] = \frac{q}{2} \times 2 q^{2} = q^{3} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If ⁿp_r = 840, ⁿC_r = 35, then $r =$ ______.

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t {}_{n}{P_{r}} = 840 a n d {}_{n}{C_{r}} = 35 \\ \Rightarrow \frac{n!}{(n - r)!} = 840 \dots (i) a n d \frac{n!}{r! (n - r)!} = 35 \dots (i i) \\ D i v i d i n g e q n . (i) b y (i i) w e g e t \\ \frac{\frac{n!}{(n - r)!}}{\frac{n!}{r! (n - r)!}} = \frac{840}{35} \Rightarrow \frac{n!}{(n - r)!} \times \frac{r! (n - r)!}{n!} = 24 \\ \Rightarrow r! = 24 \Rightarrow r! = 4 \times 3 \times 2 \times 1 \\ \Rightarrow r! = 4! ∴ r = 4 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 4 . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

If $x, 2 y, 3 z$ are in A.P., where the distinct numbers $x, y, z$ are in G.P., then the common ratio of the G.P. is

(a) 3

(b) $\frac{1}{3}$ 

(c) 2

(d) $\frac{1}{2}$

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} Since x, 2 y, 3 z a r e i n A . P . \\ ∴ 2 y - x = 3 z - 2 y \\ \Rightarrow 4 y = x + 3 z \dots (i) \\ N o w x, y, z a r e i n G . P . \\ ∴ C o m m o n r a t i o r = \frac{y}{x} = \frac{z}{y} \dots (i i) \\ ∴ y^{2} = x z \\ P u t t i n g t h e v a l u e o f x f r o m e q n . (i), w e g e t \\ y^{2} = (4 y - 3 z) z \Rightarrow y^{2} = 4 y z - 3 z^{2} \\ \Rightarrow 3 z^{2} - 4 y z + y^{2} = 0 \Rightarrow 3 z^{2} - 3 y z - y z + y^{2} = 0 \\ \Rightarrow 3 z (z - y) - y (z - y) = 0 \Rightarrow (3 z - y) (z - y) = 0 \\ \Rightarrow 3 z - y = 0 a n d z - y = 0 \\ \Rightarrow 3 z = y a n d z \neq y [? z a n d y a r e d i s t i n c t n u m b e r s] \\ \Rightarrow \frac{z}{y} = \frac{1}{3} \Rightarrow r = \frac{1}{3} [? F r o m e q n . (i i)] \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

If 9 times the 9th term of an A.P. is equal to 13 times the 13th term, then the 22nd term of the A.P. is

(a) 0

(b) 22

(c) 220

(d) 198

10 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t T_{n} = a + (n - 1) d \\ ∴ T_{9} = a + (9 - 1) d \Rightarrow T_{9} = a + 8 d \\ a n d T_{13} = a + 12 d \\ A s p e r t h e g i v e n c o n d i t i o n \\ 9 [a + 8 d] = 13 [a + 12 d] \\ \Rightarrow 9 a + 72 d = 13 a + 156 d \Rightarrow - 4 a = 84 d \\ \Rightarrow a = - 21 d \dots (i) \\ N o w, T_{22} = a + 21 d = - 21 d + 21 d = 0 [F r o m e q n . (i)] \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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