How-many-ways-are-there-to-arrange-the-letters-of-the-word-VISITING-if-no-two-I-s-are-adjacent-

Question Number 124826 by bramlexs22 last updated on 06/Dec/20

H o w m a n y w a y s a r e t h e r e t o a r r a n g e

t h e l e t t e r s o f t h e w o r d^{'} {V I S I T I N G}^{'}

i f n o t w o I^{'} s a r e a d j a c e n t ?

Answered by mr W last updated on 06/Dec/20

M e t h o d I

a r r a n g e a t f i r s t t h e f i v e o t h e r l e t t e r s

a n d t h e n t h e t h r e e I^{'} s i n 6 p o s i t i o n s :

V S T N G

5! \times C_{3}^{6} = 2400

M e t h o d I I

a r r a n g e a t f i r s t t h e t h r e e I^{'} s a n d

t h e n t h e f i v e o t h e r l e t t e r s :

I I I

= z e r o o r m o r e o t h e r l e t t e r s

= o n e o r m o r e o t h e r l e t t e r s

{(\underset{}{1 + x + x^{2} + \dots})}^{2} {(\underset{}{x + x^{2} + x^{3} + \dots})}^{2}

= \frac{x^{2}}{{(1 - x)}^{4}}

= x^{2} \underset{k = 0}{\sum^{\infty}} C_{3}^{k + 3} x^{k}

c o e f . o f t e r m x^{5} :

k = 3 \Rightarrow C_{3}^{6}

\Rightarrow C_{3}^{6} \times 5! = 2400

Commented by john_santu last updated on 06/Dec/20

g r e a t m e t h o d I I

Commented by malwan last updated on 06/Dec/20

I w a n t t o m a s t e r m e t h o d I I

l i k e y o u m r W

t h a n k y o u

Commented by mr W last updated on 06/Dec/20

{i t}^{'} s a v e r y p o w e r f u l m e t h o d, t r y t o

a p p l y i t!

Answered by john_santu last updated on 06/Dec/20

Commented by bramlexs22 last updated on 06/Dec/20

g r e a t!