Maths NCERT Exemplar Solutions Class 11th Chapter Seven

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Payal Gupta

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W e k n o w t h a t a c o i n h a s H e a d a n d T a i l ( H , T ) W h e n a c o i n i s t o s s e d 6 t i m e s , t h e n t h e P o s s i b l e o u t c o m e = 2 6 = 6 4 H e n c e , t h e c o r r e c t o p t i o n i s ( b ) .

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Payal Gupta

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G i v e n t h a t : C 1 2 n = C 8 n C n 1 2 n = C 8 n [ ? C r n = C n r n ] n 1 2 = 8 n = 8 + 1 2 = 2 0 H e n c e , t h e c o r r e c t o p t i o n i s ( a ) .

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Payal Gupta

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W e h a v e 4 g i r l s a n d 7 b o y s a n d a t e a m o f 5 m e m b e r s i s t o b e s e l e c t e d . ( i ) I f n o g i r l i s s e l e c t e d , t h e n a l l t h e 5 m e m b e r s a r e t o b e s e l e c t e d o u t o f 7 b o y s i . e . C 5 7 = 7 ! 5 ! 2 ! = 7 * 6 . 5 ! 5 ! * 2 = 2 1 w a y s ( i i ) W h e n a t l e a s t o n e b o y a n d o n e g i r l a r e t o b e s e l e c t e d , t h e n N u m b e r o f w a y s = C 1 4 * C 4 7 + C 2 4 * C 3 7 + C 3 4 * C 2 7 + C 4 4 * C 1 7 = 4 * 7 * 6 * 5 * 4 4 * 3 * 2 * 1 + 4 * 3 2 * 1 * 7 * 6 * 5 3 * 2 * 1 + 4 * 7 * 6 2 * 1 + 1 * 7 = 4 * 3 5 + 6 * 3 5 + 4 * 2 1 + 7 = 1 4 0 + 2 1 0 + 8 4 + 7 = 4 4 1 w a y s ( i i i ) W h e n a t l e a s t 3 g i r l s a r e i n c l u d e d , t h e n N u m b e r o f w a y s = C 3 4 * C 2 7 + C 4 4 * C 1 7 = 4 * 7 * 6 2 * 1 + 1 * 7 = 8 4 + 7 = 9 1 w a y s H e n c e , t h e r e q u i r e d n u m b e r o f w a y s a r e ( i ) 2 1 w a y s ( i i ) 4 4 1 w a y s ( i i i ) 9 1 w a y s

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Payal Gupta

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

T o t a l n u m b e r o f s t u d e n t s i n e a c h c l a s s = 2 0 W e h a v e t o s e l e c t a t l e a s t 5 s t u d e n t s f r o m e a c h c l a s s . W e h a v e t h e f o l l o w i n g c a s e s . ( i ) 5 s t u d e n t s f r o m X I c l a s s a n d 6 s t u d e n t s f r o m X I I c l a s s ( i i ) 6 s t u d e n t s f r o m X I c l a s s a n d 5 s t u d e n t s f r o m X I I c l a s s S o , n u m b e r o f w a y s o f s e l e c t i o n o f a t e a m o f 1 1 p l a y e r s = C 5 2 0 * C 6 2 0 + C 6 2 0 * C 5 2 0 = 2 [ C 5 2 0 * C 6 2 0 ] H e n c e , t h e r e q u i r e d w a y s o f s e l e c t i o n = 2 [ C 5 2 0 * C 6 2 0 ]

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Payal Gupta

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

G i v e n t h a t t h e t o t a l n u m b e r o f p l a y e r s = 1 6 W e h a v e t o s e l e c t 1 1 p l a y e r s o u t o f 1 6 p l a y e r s . ( i ) I f 2 p l a y e r s a r e i n c l u d e d , t h e n n u m b e r o f w a y s o f s e l e c t i o n = C 1 1 2 1 6 2 = C 9 1 4 ( i i ) I f 2 p l a y e r s a r e e x c l u d e d , t h e n n u m b e r o f w a y s o f s e l e c t i o n = C 1 1 1 6 2 = C 1 1 1 4 H e n c e , t h e r e q u i r e d n u m b e r o f w a y s o f s e l e c t i o n a r e ( i ) C 9 1 4 ( i i ) C 1 1 1 4

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Payal Gupta

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T o t a l n u m b e r o f m a r b l e s = 6 w h i t e + 5 r e d = 1 1 m a r b l e s (i)Since,wehavetodraw4marblesofanycolourfromthe11marbles Requirednumberofways=C411 ( i i ) I f 2 m u s t b e w h i t e a n d 2 m u s t b e r e d , t h e n t h e r e q u i r e d n u m b e r o f w a y s = C 2 6 * C 2 5 ( i i i ) I f a l l t h e 4 m a r b l e s a r e o f t h e s a m e c o l o u r , t h e n t h e r e q u i r e d n u m b e r o f w a y s = C 4 6 + C 4 5 H e n c e , t h e r e q u i r e d n u m b e r o f w a y s a r e ( i ) C 4 1 1 ( i i ) C 2 6 * C 2 5 ( i i i ) C 4 6 + C 4 5

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Payal Gupta

Contributor-Level 10

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

Giventhat18micewereplacedequallyintwoexperimentalgroupsand o n e c o n t r o l g r o u p i . e . 3 g r o u p s T h e r e q u i r e d n u m b e r o f a r r a n g e m e n t s = T o t a l a r r a n g e m e n t s E q u a l l y l i k e l y a r r a n g e m e n t s = 1 8 ! 6 ! 6 ! 6 ! = 1 8 ! ( 6 ! ) 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 = 1 8 ! ( 6 ! ) 3

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Payal Gupta

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

  G i v e n t h a t q u e s t i o n n u m b e r 1 a n d 2 a r e c o m p u l s o r y . T h e r e m a i n i n g q u e s t i o n s a r e 5 2 = 3 L e t n b e t h e n u m b e r o f s i d e s i n a p o l y g o n . Since,Polygonofnsideshas(C2nn)numberofdiagonals C 2 n n = 4 4 = n ! 2 ! ( n 2 ) ! n = 4 4 n ( n 1 ) ( n 2 ) ! 2 . ( n 2 ) ! n = 4 4 n ( n 1 ) 2 n = 4 4 n 2 n 2 n 2 = 4 4 n 2 3 n = 8 8 n 2 3 n 8 8 = 0 n 2 1 1 n + 8 n 8 8 = 0 n ( n 1 1 ) + 8 ( n 1 1 ) = 0 ( n 1 1 ) ( n + 8 ) = 0 n = 1 1 a n d n = 8 [ ? n 8 ] S o , n = 1 1 H e n c e , t h e r e q u i r e d n u m b e r o f s i d e s = 1 1 .

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Payal Gupta

Contributor-Level 10

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

  G i v e n t h a t q u e s t i o n n u m b e r 1 a n d 2 a r e c o m p u l s o r y . T h e r e m a i n i n g q u e s t i o n s a r e 5 2 = 3 T o t a l n u m b e r o f q u e s t i o n s t o b e a t t e m p t e d = 4 q u e s t i o n s 1 a n d 2 a r e c o m p u l s o r y . S o o n l y 2 q u e s t i o n s a r e t o b e d o n e o u t o f 3 q u e s t i o n s T h e r e f o r e n u m b e r o f w a y s = C 2 3 = C 3 2 3 = C 1 3 = 3 [ ? C r n = C n r n ] H e n c e , t h e r e q u i r e d n u m b e r o f w a y s = 3 .

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Payal Gupta

Contributor-Level 10

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

  I f f i r s t t w o d i g i t s i s 4 1 , t h e n t h e r e m a i n i n g 4 d i g i t s c a n b e a r r a n g e d i n P 4 8 w a y s = 8 ! ( 8 4 ) ! = 8 ! 4 ! = 8 * 7 * 6 * 5 * 4 ! 4 ! = 1 6 8 0 S i m i l a r l y f i r s t t w o d i g i t s c a n b e 4 2 o r 4 6 o r 6 2 o r 6 4 . T o t a l n u m b e r o f t e l e p h o n e n u m b e r s h a v e a l l d i g i t s d i s t i n c t = 5 * 1 6 8 0 = 8 4 0 0 H e n c e , t h e r e q u i r e d t e l e p h o n e n u m b e r s = 8 4 0 0 .

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