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If the letters of the word ASSASSINATION are arranged at random, find the probability that:

(a) Four S’s come consecutively in the word.

(b) Two I’s and two N’s come together.

(c) All A’s are not coming together.

(d) No two A’s are coming together.

0 3 Views | Posted 3 months ago
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    Answered by

    Payal Gupta | Contributor-Level 10

    3 months ago

    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     w o r d     i s     A S S A S S I N A T I O N     a r e     1 3 . W h e r e ,     w e     h a v e     3 A ' s ,     4 S ' s ,     2 I ' s ,     2 N ' s ,     1 T ' s     a n d     1 O ' s . (a)  If  4S's  come  consecutively  in  the  word,  then  arrangement  may  be  as  follows:                                   S S S S 1     G r o u p     A A A I I N N T O ? 9     o t h e r s ∴           N u m b e r     o f     w o r d s     w h e n     a l l     S ' s     a r e     t o g e t h e r = 1 0 ! 3 ! 2 ! 2 !

                  a n d     t h e     t o t a l     n u m b e r     o f     w o r d     f o r m e d     f r o m     t h e     w o r d s     A S S A S S I N A T I O N = 1 3 ! 3 ! 4 ! 2 ! 2 ! ∴           R e q u i r e d     p r o b a b i l i t y = 1 0 ! 3 ! 2 ! 2 ! 1 3 ! 3 ! 4 ! 2 ! 2 ! = 1 0 ! 3 ! 2 ! 2 ! × 3 ! 4 ! 2 ! 2 ! 1 3 !                                                                                                                 = 1 0 ! 4 ! 1 3 ! = 1 0 !   × 4 × 3 × 2 1 3 × 1 2 × 1 1 × 1 0 ! = 2 1 4 3 ( b )     I f     2 I ' s     a n d     2 N ' s     c o m e     t o g e t h e r     t h e n     t h e r e     a r e     1 0     a l p h a b e t s                 N u m b e r     o f     w o r d s     w h e n     2 I ' s     a n d     2 N ' s     c o m e     t o g e t h e r = 1 0 ! 3 ! 4 ! × 4 ! 2 ! 2 ! ∴           R e q u i r e d     p r o b a b i l i t y = 1 0 ! 3 ! 4 ! × 4 ! 2 ! 2 ! 1 3 ! 3 ! 4 ! 2 ! 2 ! = 1 0 ! 3 ! 4 ! × 4 ! 2 ! 2 ! × 3 ! 4 ! 2 ! 2 ! 1 3 !                                                                                                                 = 4 ! 1 0 ! 1 3 ! = 4 × 3 × 2 × 1 0 ! 1 3 × 1 2 × 1 1 × 1 0 ! = 2 1 4 3 (c)  If  all  A's  are  coming  together,  then  there  are  11  alphabets               N u m b e r     o f     w o r d s     w h e n     a l l     A ' s     c o m e     t o g e t h e r = 1 1 ! 4 ! 2 ! 2 ! ∴           P r o b a b i l i t y     w h e n     a l l     A ' s     c o m e     t o g e t h e r = 1 1 ! 4 ! 2 ! 2 ! 1 3 ! 4 ! 3 ! 2 ! 2 ! = 1 1 ! 4 ! 2 ! 2 ! × 4 ! 3 ! 2 ! 2 ! 1 3 !                                                                                                                                                                                                   = 1 1 ! 3 ! 1 3 ! = 6 1 3 × 1 2 = 1 2 6 ∴       R e q u i r e d     p r o b a b i l i t y     w h e n     a l l     A ' s     d o     n o t     c o m e     t o g e t h e r = 1 − 1 2 6 = 2 5 2 6 ( d )     I f     n o     t w o     A ' s     a r e     t o g e t h e r ,     t h e n     a r r a n g i n g     t h e     a l p h a b e t s     e x c e p t     A ' s                   − S − S − S − S − I − N − T − I − O − N −                   N u m b e r     o f     w a y s     o f     a r r a n g i n g     a l l     a l p h a b e t s       e x c e p t     A ' s = 1 0 ! 4 ! 2 ! 2 !                   T h e r e     a r e     1 1     v a c a n t     p l a c e s     b e t w e e n     t h e s e     a l p h a b e t s . ∴               3 A ' s     c a n     b e     p l a c e d     i n     1 1     p l a c e s     i n     C 3 1 1     w a y s = 1 1 ! 3 ! 8 ! ∴             T o t a l     n u m b e r     o f     w o r d s     w h e n     n o     t w o     A ' s     c o m e     t o g e t h e r = 1 1 ! 3 ! 8 ! × 1 0 ! 4 ! 2 ! 2 ! ∴           R e q u i r e d     p r o b a b i l i t y = 1 1 ! 3 ! 8 ! × 1 0 ! 4 ! 2 ! 2 ! × 4 ! 3 ! 2 ! 2 ! 1 3 !                                                                                                               = 1 0 ! 8 !   × 1 3 × 1 2 = 1 0 × 9 × 8 ! 8 !   × 1 3 × 1 2 = 1 0 × 9 1 3 × 1 2 = 1 5 2 6

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