**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.

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Payal Gupta | Contributor-Level 10

8 months ago

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

$\begin{array}{l} T o t a l n u m b e r o f w o r d i s A S S A S S I N A T I O N a r e 13 . \\ W h e r e, w e h a v e 3 A' s, 4 S' s, 2 I' s, 2 N' s, 1 T' s a n d 1 O' s . \\ (a) I f 4 S' s c o m e consecutively i n t h e w o r d, t h e n a r r a n g e m e n t m a y b e a s f o l l o w s : \\ \underset{1 G r o u p}{S S S S} \underset{9 o t h e r s}{\underset{?}{A A A I I N N T O}} \\ ∴ N u m b e r o f w o r d s w h e n a l l S' s a r e t o g e t h e r = \frac{10!}{3! 2! 2!} \end{array}$

$\begin{array}{l} 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 = \frac{13!}{3! 4! 2! 2!} \\ ∴ R e q u i r e d p r o b a b i l i t y = \frac{\frac{10!}{3! 2! 2!}}{\frac{13!}{3! 4! 2! 2!}} = \frac{10!}{3! 2! 2!} * \frac{3! 4! 2! 2!}{13!} \\ = \frac{10! 4!}{13!} = \frac{10! * 4 * 3 * 2}{13 * 12 * 11 * 10!} = \frac{2}{143} \\ (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 10 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 = \frac{10!}{3! 4!} * \frac{4!}{2! 2!} \\ ∴ R e q u i r e d p r o b a b i l i t y = \frac{\frac{10!}{3! 4!} * \frac{4!}{2! 2!}}{\frac{13!}{3! 4! 2! 2!}} = \frac{10!}{3! 4!} * \frac{4!}{2! 2!} * \frac{3! 4! 2! 2!}{13!} \\ = \frac{4! 10!}{13!} = \frac{4 * 3 * 2 * 10!}{13 * 12 * 11 * 10!} = \frac{2}{143} \\ (c) I f a l l A' s a r e coming t o g e t h e r, t h e n t h e r e a r e 11 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 a l l A' s c o m e t o g e t h e r = \frac{11!}{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 = \frac{\frac{11!}{4! 2! 2!}}{\frac{13!}{4! 3! 2! 2!}} = \frac{11!}{4! 2! 2!} * \frac{4! 3! 2! 2!}{13!} \\ = \frac{11! 3!}{13!} = \frac{6}{13 * 12} = \frac{1}{26} \\ ∴ 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 - \frac{1}{26} = \frac{25}{26} \\ (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 = \frac{10!}{4! 2! 2!} \\ T h e r e a r e 11 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 11 p l a c e s i n {}_{11}{C_{3}} w a y s = \frac{11!}{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 = \frac{11!}{3! 8!} * \frac{10!}{4! 2! 2!} \\ ∴ R e q u i r e d p r o b a b i l i t y = \frac{11!}{3! 8!} * \frac{10!}{4! 2! 2!} * \frac{4! 3! 2! 2!}{13!} \\ = \frac{10!}{8! * 13 * 12} = \frac{10 * 9 * 8!}{8! * 13 * 12} = \frac{10 * 9}{13 * 12} = \frac{15}{26} \end{array}$

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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.

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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.