Ques-8-Find-the-signum-sign-or-sgn-of-the-permutation-12345678-Hint-for-any-permutation-take-sgn-1-if-is-odd-1-if-is-even-Ques-9-Prove-that

Question Number 193871 by Mastermind last updated on 21/Jun/23

Ques . 8

Find the signum (sign or sgn) of the

permutation θ = (12345678) .

Hint : for any permutation β, take

sgn β = {_{- 1 if β is odd}^{1 if β is even}

Ques . 9

Prove that ∣ S_{n} ∣ = n! .

Ques . 10

Provd that for b \in S_{n}, sgn b = sgn b^{- 1} .

Commented by talminator2856792 last updated on 23/Jun/23

what is source of these questions .

or book name .

Answered by Rajpurohith last updated on 22/Jun/23

(8) C l e a r l y θ = (18) (17) (16) (15) (14) (13) (12)

w h i c h s h o w s θ i s a n o d d p e r m u t a t i o n .

h e n c e s g n (θ) = - 1 .

(9) S_{n} d e n o t e s t h e s e t o f a l l p e r m u t a t i o n s o f

a s e t o f n e l e m e n t s s a y A = {1, 2, 3, \dots n} .

[y o u k n o w t h a t S_{n} f o r m s

a g r o u p u n d e r f u n c t i o n c o m p o s i t i o n .]

l e t σ \in S_{n}, s o σ i s a b i j e c t i o n .

H o w m a n y c h o i c e s d o y o u h a v e f o r σ (1) ?

Y o u h a v e n c h o i c e s .

A n d f o r σ (2) ?

I t m u s t b e (n - 1) c h o i c e s .

b e c a u s e y o u c a n n o t c h o o s e t h e o n e y o u c h o o s e d

f o r σ (1) .

[S u p p o s e y o u c h o o s e σ (2) = σ (1),

\Rightarrow 2 = 1 a s σ i s b i j e c t i v e .]

s o y o u c a n n o t .

s i m i l a r l y t h e r e a r e (n - 2) c h o i c e s f o r σ (3) .

.

.

.

T h e r e a r e 2 c h o i c e s f o r σ (n - 1)

A n d o n l y o n e c h o i c e f o r σ (n) .

B y f u n d a m e n t a l p r i n c i p l e o f c o u n t i n g,

w e h a v e n (n - 1) (n - 2) \dots 2.1 = n! s u c h p e r m u t a t i o n s

o n n l e t t e r s a n d t h e y a r e t h e o n l y p e r m u t a t i o n s .

\Rightarrow∣ S_{n} ∣= n!

(10) L e t β \in S_{n} . S u p p o s e β i s e v e n .

\Rightarrow β o β^{- 1} = I d

\Rightarrow s g n (β o β^{- 1}) = s g n (I d) = 1

S u p p o s e β^{- 1} i s o d d

\Rightarrow β o β^{- 1} i s o d d \Rightarrow s g n (β o β^{- 1}) = - 1, a c o n t r a d i c t i o n .

H e n c e β^{- 1} i s n o t o d d

\Rightarrow β^{- 1} m u s t b e e v e n .

\Rightarrow s g n (β^{- 1}) = 1 = s g n (β)

S i m i l a r a r g u m e n t f o r o d d .

Commented by Mastermind last updated on 22/Jun/23

Thank you so much

Commented by York12 last updated on 22/Jun/23

s i r I a m f a c i n g a l o t o f p r o b l e m s w i t h c o m b i n a t o r i c s

w h a t d o y o u r e c o m m e n d t o r e a d

Commented by Rajpurohith last updated on 22/Jun/23

I w o u l d b e t t e r s u g g e s t y o u t o s t u d y

o n e b o o k p r o p e r l y a s a c o u r s e .

Commented by talminator2856792 last updated on 23/Jun/23

what book do you recommend .

Commented by York12 last updated on 23/Jun/23

w h a t b o o k d o y o u r e c o m m e n d s i r

Commented by Rajpurohith last updated on 24/Jun/23

A c t u a l l y I d i d n o t h a v e a c o u r s e i n i t,

B u t I^{'} d s u g g e s t t h e b o o k “ I n t r o d u c t o r y

{C o m b i n a t o r i c s}^{″} b y R i c h a r d . A B r u a l d i .

Commented by York12 last updated on 23/Jul/23

t h a n k s

t h a n k s

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