8-digit-numbers-are-formed-using-1-1-2-2-2-3-4-4-The-number-of-such-numbers-in-which-the-odd-digits-do-not-occupy-odd-places-is-a-160-b-120-c-60-d-48-

Question Number 52278 by Necxx last updated on 05/Jan/19

8 d i g i t n u m b e r s a r e f o r m e d u s i n g

1 1 2 2 2 3 4 4 . T h e n u m b e r o f s u c h

n u m b e r s i n w h i c h t h e o d d d i g i t s

d o n o t o c c u p y o d d p l a c e s i s

a) 1 60 b) 120 c) 60 d) 48

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jan/19

\underset{8}{-} \underset{7}{-} \underset{6}{-} \underset{5}{-} \underset{4}{-} \underset{3}{-} \underset{2}{-} \underset{1}{-}

o d d p l a c e a r e (1, 3, 5, 7)

e v e n p l a c e (2, 4, 6, 8)

o d d d i g i t (1, 1, 3) \to t h r e e

e v e n d i g i t (2, 2, 2, 4, 4) \to f i v e

n o w

p l a c e 1, 3, 5, 7 t o b e f i l l e d b y e v e n d i g i t s (2, 2, 2, 4, 4)

s o n u m b e r s o f w a y i s = \frac{5 \times 4 \times 3 \times 2}{3! \times 2!} = 10

[d e n o m i n a t o r 3! f o r (2, 2, 2) a n d 2! f o r (4, 4)]

r e m a n i n g f o u r p l a c e (2, 4, 6, 8)

r r m a i n i n g d i g i t (1, 1, 3) n d o n e e v e n d i g i t

w a y s = \frac{4 \times 3 \times 2 \times 1}{2!} = 12

o u t o f t h e s e (10 \times 12) = 120

Commented by mr W last updated on 06/Jan/19

w h a t i s y o u r f i n a l r e s u l t s i r ?

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Jan/19

t h a n k y o u s i r \dots c l u e i s i n y o u r a n s w e r \dots

Commented by mr W last updated on 06/Jan/19

t h a n k s f o r c o n f i r m i n g s i r!

Commented by Necxx last updated on 06/Jan/19

I j u s t l o v e t h i s e x p l a n a t i o n . T h a n k

y o u s i r T a n m a y .

Answered by mr W last updated on 06/Jan/19

t h e o n l y r e s t r i c t i o n : o d d d i g i t s {d o n}^{'} t o c c u p y

o d d p l a c e s, i . e . t h e t h r e e d i g i t s 1, 1, 3

c a n o n l y b e p l a c e d i n 4 e v e n p o s i t i o n s .

t o a r r a n g e t h e t h r e e o d d d i g i t s (1, 1, 3)

i n 4 e v e n p o s i t i o n s t h e r e a r e

\frac{P_{3}^{4}}{2!} d i f f e r e n t w a y s .

t o a r r a n g e t h e r e m a i n i n g 5 e v e n

d i g i t s (2, 2, 2, 4, 4) i n t h e r e m a i n i n g 5

p o s s i t i o n s t h e r e a r e

\frac{5!}{2! 3!} d i f f e r e n t w a y s .

s o t h e r e s u l t i s \frac{P_{3}^{4}}{2!} \times \frac{5!}{2! 3!} = \frac{4 \times 3 \times 2}{2} \times \frac{5 \times 4}{2} = 120

\Rightarrow a n s w e r b) i s c o r r e c t

Commented by Necxx last updated on 06/Jan/19

y e a h \dots . v e r y e x p l a n a t o r y . T h a n k s

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