\magnification 2000
\parskip 10pt
\parindent 0 pt

Show $\leq$ is a partial order:

Claim:  $\leq$ is reflexive (ie  $\pi \leq  \pi$).

Let $A$ be the set of inversions of $\pi$.  Then $A \subset A$.  Hence  $\pi \leq  \pi$. 


Claim:  $\leq$ is transitive (ie  if $\pi \leq  \sigma$  and $\sigma \leq  \tau$, then $\pi \leq \tau$).

Let $A$ be the set of inversions of $\pi$.  

Let $B$ be the set of inversions of $\sigma$. 

Let $C$ be the set of inversions of $\tau$. 

$\pi \leq  \sigma$  implies $A \subset B$.

$\sigma \leq  \tau$ implies $B \subset C$.

$A \subset B$. and $B \subset C$ implies $A \subset C$.  Hence $\pi \leq \tau$.


Claim:  $\leq$ is antisymmetric (ie  if $\pi \leq  \sigma$  and $\sigma \leq  \pi$, then $\pi = \sigma$).

Let $A$ be the set of inversions of $\pi$.  

Let $B$ be the set of inversions of $\sigma$. 

$\pi \leq  \sigma$  implies $A \subset B$.

$\sigma \leq  \pi$ implies $B \subset A$.

$A \subset B$. and $B \subset A$ implies $A  =  B$.  


Let $a_i = |\{ j ~|~ (j, i) \in A \}|$.  Then $a_1, ..., a_n$ is the inversion sequence for the permutation $\pi$.

Since $A = B$, $\pi$ and $\sigma$ have the same inversion sequence.  Hence $\pi = \sigma$ by thm 4.2.1.

\eject



5.3

Let $r \in {\cal R}$, $k \in {\cal Z}$.

Define $\left(\matrix{r \cr k}\right) = \cases {
{r(r-1) ... (r - k + 1) \over k!}  & if $k \geq 1$  \cr
1 & if  $k = 0$ \cr
0 & if $k \leq -1$ \cr
} $

\vskip 10pt
5.6:  Newton's Binomial Theorem

Thm 5.6.1:  Let $\alpha \in {\cal R}$.  Then if $ 0 \leq |x| < |y|$,
$$(x + y)^\alpha = \Sigma_{k = 0}^\infty  \left(\matrix{\alpha \cr k}\right)  x^k y^{\alpha - k}$$
\vfill
\eject

Thm 5.4.1:  Let $n$ be a positive integer.  The sequence of binomial coefficients is a unimodal sequence.  In particular if $n$ is even,

$$ \left(\matrix{n \cr 0}\right)  < \left(\matrix{n \cr 1}\right)  ...  < \left(\matrix{n \cr {n \over 2}}\right) $$


$$ \left(\matrix{n \cr {n \over 2}}\right) > ...  >  \left(\matrix{n \cr n-1}\right)   > \left(\matrix{n \cr n}\right) $$

and if $n$ is odd

$$ \left(\matrix{n \cr 0}\right)  < \left(\matrix{n \cr 1}\right)  ...  < \left(\matrix{n \cr {n -1 \over 2}}\right) =  \left(\matrix{n \cr {n +1 \over 2}}\right)  $$


$$ \left(\matrix{n \cr {n + 1 \over 2}}\right) > ...  >  \left(\matrix{n \cr n-1}\right)   > \left(\matrix{n \cr n}\right) $$


Proof idea:  Look at ${ \left(\matrix{n \cr k }\right)  \over  \left(\matrix{n \cr k - 1 }\right) } = {n - k + 1 \over k}  $
\eject

5.5:  Multinomial thm

Define $\left(\matrix{n \cr n_1 n_2 ... n_t }\right)   = {n! \over n_1! n_2! ... n_t!} $

Thm 5.5.1:  Let $n \in {\cal Z}$.  Then
$$(x_1 + x_2 + ... x_t)^n = \Sigma \left(\matrix{n \cr n_1 n_2 ... n_t }\right)  x_1^{n_1} x_2^{n_2} + ... x_t^{n_t}$$

where the summation extends over all nonnegative integral solutions to $n_1 + n_2 + ... + n_t  = n$


\end
























4321:  $\{  (4, 3), (4, 2), (4, 1), (3, 2), (3, 1), (2, 1) \}$


4312:  $\{ (4, 3), (4, 1), (4, 2), (3, 1), (3, 2) \}$
\hfil
3421:  $\{  (4, 2), (4, 1), (3, 2), (3, 1), (2, 1) \}$
\hfil
4231:  $\{ (2, 1), (3, 1), (4, 1), (4, 2), (4, 3) \}$


4132: $\{  (4,1), (4, 2), (4, 3), (3,2) \}$
\hfil
3412:   $\{ (3, 1),  (3,2), (4, 1), (4, 2) \}$
\hfil
3241:  $\{ (4, 1), (3, 2), (3, 1), (2, 1), \}$
\hfil
2431:  $\{ (2, 1), (3, 1), (4, 1), (4, 3) \}$
\hfil
4213:  $\{ (2, 1),  (4, 1), (4, 2), (4, 3) \}$



4123: $\{  (4,1), (4, 2), (4, 3) \}$
\hfil
1432:   $\{  (4, 2), (4, 3), (3,2) \}$
\hfil
3142:   $\{ (3, 1),  (3,2), (4, 2) \}$
\hfil
3214:  $\{ (3, 2), (3, 1), (2, 1) \}$
\hfil
2341:  $\{ (2, 1), (3, 1), (4, 1) \}$  
\hfil
2413:  $\{ (2, 1),  (4, 1),  (4, 3) \}$


1423: $\{  (4, 2), (4, 3) \}$
\hfill
1342:   $\{ (4, 2), (3,2) \}$
\hfill
3124:   $\{ (3, 1),  (3,2) \}$
\hfill
2314:  $\{ (2, 1), (3, 1) \}$
\hfill
2143:  $\{ (2, 1),   (4, 3) \}$