In-how-many-ways-can-the-word-KINECTIC-be-arranged-so-that-no-vowels-can-be-together-

Question Number 206939 by Tawa11 last updated on 01/May/24

In how many ways can the word KINECTIC be
arranged so that no vowels can be together?

Commented by mr W last updated on 01/May/24

3600 ?

Commented by Tawa11 last updated on 01/May/24

I got same answer sir .

Commented by Tawa11 last updated on 01/May/24

But for KINETIC

I got 2160

Textbook says 720

Commented by A5T last updated on 01/May/24

I g u e s s t h e t e x t b o o k i s r i g h t a b o u t 720 .

Commented by mr W last updated on 01/May/24

t h e q u e s t i o n s a i d K I N E C T I C, n o t

K I N E T I C .

i f K I N E C T I C i s m e a n t, t h e n a n s w e r

i s 3 \times 20 \times \frac{5!}{2} = 3600 .

i f K I N E T I C i s m e a n t, t h e n a n s w e r

i s 3 \times 10 \times 4! = 720 .

Commented by BaliramKumar last updated on 01/May/24

solutio \underset{}{n}

Commented by Tawa11 last updated on 01/May/24

Thanks sir .

I understand now sir .

Answered by mr W last updated on 01/May/24

K I N E C T I C

t h e v o w e l s (I, I, E) m u s t b e s e p a r a t e d

b y a t l e a s t o n e c o n s o n a n t .

X V Y V Y V X

V = o n e v o w e l

X = p l a c e h o l d e r f o r z e r o o r m o r e c o n s o n a n t s

Y = p l a c e h o l d e r f o r o n e o r m o r e c o n s o n a n t s

t o p l a c e t h e v o w e l s t h e r e a r e \frac{3!}{2!} w a y s .

t o p l a c e t h e 5 c o n s o n a n t s t h e r e

a r e f o l l o w i n g p o s s i b i l i t i e s :

0 + 1 + 1 + 3

0 + 1 + 2 + 2

0 + 1 + 3 + 1

0 + 1 + 4 + 0

0 + 2 + 1 + 2

0 + 2 + 2 + 1

0 + 2 + 3 + 0

0 + 3 + 1 + 1

0 + 3 + 2 + 0

0 + 4 + 1 + 0

1 + 1 + 1 + 2

1 + 1 + 2 + 1

1 + 1 + 3 + 0

1 + 2 + 1 + 1

1 + 2 + 2 + 0

1 + 3 + 1 + 0

2 + 1 + 1 + 1

2 + 1 + 2 + 0

2 + 2 + 1 + 0

3 + 1 + 1 + 0

{t h a t}^{'} s t o t a l l y 20 p o s s i b i l i t i e s .

t h e r e f o r e t h e r e a r e

\frac{3!}{2!} \times 20 \times \frac{5!}{2!} = 3600 w a y s

s i m i l a r l y f o r K I N E T I C t h e r e a r e

\frac{3!}{2!} \times 10 \times 4! = 720 w a y s

Commented by mr W last updated on 02/May/24

M e t h o d I I (g e n e r a t i n g f u n c t i o n)

f o r w o r d K I N E C T I C

X V Y V Y V X

X \Rightarrow (1 + x + x^{2} + \dots .)

Y \Rightarrow (x + x^{2} + \dots .) = x (1 + x + x^{2} + \dots)

{(1 + x + x^{2} + \dots .)}^{2} {(x + x^{2} + \dots .)}^{2}

= x^{2} {(1 + x + x^{2} + \dots .)}^{4}

= \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 x^{5} t e r m i s C_{3}^{3 + 3} = 20

\Rightarrow n u m b e r o f w a y s : \frac{3!}{2!} \times 20 \times \frac{5!}{2!} = 3600

f o r w o r d K I N E T I C

c o e f . o f x^{4} t e r m i s C_{3}^{2 + 3} = 10

\Rightarrow n u m b e r o f w a y s : \frac{3!}{2!} \times 10 \times 4! = 720

Commented by Tawa11 last updated on 01/May/24

Great sir .

I will study this .

Commented by mr W last updated on 02/May/24

g e n e r a l l y i f t h e r e a r e m (d i f f e r e n t)

v o w e l s a n d n (d i f f e r e n t) c o n s o n a n t s

(n + 1 ⩾ m), t h e n t h e a n s w e r i s

m! n! C_{m}^{n + 1}

i f t h e r e i s r e p e t i t i o n i n v o w e l s a n d

o r i n c o n s o n a n t s, t h e n t h e a n s w e r i s

\frac{m!}{m_{1}! m_{2}! \dots} \times \frac{n!}{n_{1}! n_{2}! \dots} \times C_{m}^{n + 1}

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