be real numbers.Prove that
be a positive integer, and consider an 
grid. A right-down path is a sequence of grid cells such that each cell is either one cell to the right of or one cell below the previous cell in the sequence. A right-up path is a sequence of grid cells such that each cell is either one cell to the right of or one cell above the previous cell in the sequence.
grid uses 
paths.
![\[f(xy)+f(xf(x^2y))=f(x^2)+f(y^2)+f(f(xy^2))+x \]](http://latex.artofproblemsolving.com/d/7/b/d7bf030b047d49477eba887e8bb78c6e5b126fe7.png)
holds for all 
.
such that 
.
Where 
be a positive semi-definite matrix. Prove that the matrix 
,
.
?
such that 
for all positive integers 
and 
be two points that are diametrically opposite to each other on a unit sphere. 
,
,
,
and 
respectively, and the points 
are collinear.
where 
is the inverse gamma function, and 
is the inverse of 
is a polynomial with integer coefficients, with 
odd. Suppose that 
for all 
Prove that 
is nonzero for all 
,
Prove that 
.
converges then so does 
for integers,
= 
= 
, as 
the 
which gets us the 
form therefore L'hopital came to my mind , which might be a completely wrong intuition, anyway what should i do to find that limit
Y by Davi-8191, yojan_sushi, opptoinfinity, megarnie, ImSh95, Adventure10, Mango247, Tellocan, Deadline, sami1618, cubres, and 10 other users
Let 
be an integer. We are given a balance and 
weights of weight 
. We are to place each of the weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step we choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all of the weights have been placed.
Determine the number of ways in which this can be done.
Proposed by Morteza Saghafian, Iran
be an integer. We are given a balance and 
weights of weight 
. We are to place each of the weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step we choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all of the weights have been placed.Determine the number of ways in which this can be done.
Proposed by Morteza Saghafian, Iran
of creating a sum 
,
,
is a permutation of 
;
,
is always different than zero.
,
.
)
cannot be at the begining, but otherwise, 
[/color]$](http://latex.artofproblemsolving.com/8/8/3/88375cf3ccf8db5aa8edcbba5b903b9f9a37ba02.png)
.
,
.
which is the product of the odd numbers from 
for 
,
.
,
is a permutation of 
,
for every 
(notice that a signed sum of different powers of 
,
.
be the number which we have to find. Let 
be last placed weight. Number of ways that right pan is never heavier than left pan placing by 
is 
(It's not hard to see).
placing 
, there is 
ways to do it , obviously.
and 
, which implies 
or the weights 
.
with 
we get the same number 
.
possiblie ways for the 
weights that have been placed before. In the last steps we can place the weights in any order on any pan, which gives 
ways.
and 
.
.
be the number of ways to choose k numbers 
from 
and distributing the weights 
in the manner described in the statement of the problem. By definition, 
.
and then picking a position 
(after which, the weights can be distributed randomly), we get the recursion ![\[a(n,k) = \sum_{g=k}^n \sum_{p=1}^k a(g-1,p-1)\binom{g-p}{k-p}(k-p)!2^{k-p}.\]](http://latex.artofproblemsolving.com/a/e/6/ae67747bbce2405dcab5e31c859e16694cabb3ff.png)
we get 
,
,
,
for some 
.
means we found ![\[\begin{aligned} c_k\binom{n}{k} & = \sum_{g=k}^n \sum_{p=1}^k c_{p-1}\binom{g-1}{p-1}\binom{g-p}{k-p}(k-p)!2^{k-p} \\ & = \sum_{g=k}^n \sum_{p=1}^k c_{p-1}\frac{(g-1)!}{(p-1)!(g-k)!}2^{k-p} \\ & = \left(\sum_{g=k}^n \frac{(g-1)!}{(g-k)!}\right)\left(\sum_{p=1}^k \frac{c_{p-1}2^{k-p}}{(p-1)!}\right).\end{aligned}\]](http://latex.artofproblemsolving.com/6/a/7/6a729c3f7a10d2a5896ad9c413e1d31915b8f9a8.png)
.
,![\[k\alpha_k = \sum_{p=1}^k \alpha_{p-1} 2^{k-p}.\]](http://latex.artofproblemsolving.com/5/e/a/5eabad93ef7e8442b6a34ac807c16e2494a25a81.png)
be the generating function of the sequence 
.
in the expansion of 
,![\[f(X)\frac{1}{1-2X} = \sum_{k\geq 0} k\alpha_kX^{k-1} = f'(X).\]](http://latex.artofproblemsolving.com/0/a/7/0a7b4478fedc5b2109fc363def0b319e4e8473c1.png)
Solving this equation, and using 
,![\[f(X) = \frac{1}{\sqrt{1-2X}},\]](http://latex.artofproblemsolving.com/c/d/e/cdea33bbe55144994f86d057a719a0e05c4ac25b.png)
and we have, 
![\[f^{(k)}(X) = (2k-1)!! \frac{1}{\sqrt{(1-2X)^{2k+1}}},\]](http://latex.artofproblemsolving.com/c/1/d/c1d40770ea7689610721be88e42120544107b5bd.png)
so we get 
.
.
only allows for 
ways to be placed, so there is more than just considering the smallest weight.
has the property that 
for all 
.
value. The last weight in a full series of 
would have been placed in 
number of ways, no matter the value of 
must be placed on the left pan, while otherwise can be placed on either pan (since 
has been already placed - on the left pan). This amounts to 
ways, so 
.
(or even 
,
.
= 
+
+ ... + 
in the balance equals to that of 
.
,
denotes the double factorial.
denote the number of valid arrangements satisfying the initial problem. If we consider the 
weights 
and divide each by 
.
into 
places, however it cannot be placed on the right scale on the first move. Therefore there must be precisely 
valid places where 
.
.
.
that this claim holds. Then ![\[f(n) = (2n - 1)f(n - 1) = (2n - 1)(2n - 3)!! = (2n - 1)!!,\]](http://latex.artofproblemsolving.com/f/b/2/fb2e107a4a12b3967212fa28d169e2402f00ff15.png)
and the induction is complete. 
by considering the largest weight.
,
.
as the coefficient of 
in 
where 
for 
.
,
therefore, 
.
,
and therefore 
.
to be the number of ways to place 
weights on the scale. Say for a given sequence, weight 
weights that weigh less, and 
that weigh more. For the 
ways. The remaining 
ways. Finally, the 
ways. We therefore have 
This gives the number of ways to place the weights when weight 
,
.
weights, and weight 
is placed first, we have: 
ways of doing so. This is 
,
denote the number of ways to place the weights 
on the balance.
To prove this, perform casework on when the 
is trivial to see.
(in other words, 
ways to do this. Now, notice that there are 
ways to place these 
ways to determine the order of the remaining weights and 
ways to determine each weight's side on the balance. Putting it all together, there are
ways to place the weights if the largest weight is placed 
with induction. The base case, 
Writing 
as 
,
However, from the inductive hypothesis, we know that 
,
.
ways to place the smallest weight, and 
.
ways to place the weights.![\[\prod_{k=0}^{n-1} 2k + 1 = (2n = 1)!!\]](http://latex.artofproblemsolving.com/4/3/0/43081ef67d1be8fe70b91afdd3e78e4bce6a3a64.png)
so we are done.
denote the number of ways to place 
is clearly even, and if it is positive (as we require) then it is at least 2. Thus, adding one will still ensure this difference stays positive. Thus, it can be placed in 
since these weights 
are simply the previous weights 
scaled up by 2. Thus,![\[N(n)=(2n-1)N(n-1)\]](http://latex.artofproblemsolving.com/c/f/8/cf8b9b6e2fa7ff74eb7762255beb02947e6c93ee.png)
Clearly, 
(just place the only weight on the left pan). Thus, by a simple induction we obtain that![\[N(n)=(2n-1)!!\]](http://latex.artofproblemsolving.com/c/d/7/cd747b90f09404532f4d0335ac3956e05b8d4bc5.png)
for all positive integers 
,
as the number of ways to place weights weighing 
on the scale subject to the conditions of the problem. Observe that 
,
on the left pan it does not matter how we place the rest. From here, it is not hard to find the recursion
where 
runs through all subsets of 
The latter subset corresponds to the weights we place on the pan before the weight weighing 
,
has a common value.
is vacuously true (Common value is 
no matter what 
.
yield a common value for 
is the common value of all 
From here we can extract the general form 
.
,
is placed
before the 
ways. Then multiply by 
because for the rest of the items anything can happen.
for all values of 
for positive 
to 
spots) and an extra ![\[\prod_{i = 0}^{n-1} (2n-2i-1) = \boxed{(2n-1)!!}\]](http://latex.artofproblemsolving.com/5/9/e/59ef9dbfb01f7aae224d0a5baef70324701ea57c.png)
where we used a double factorial.
)
.
t
to select and permute the elements after it, 
ways to assign them to the balance. For the weights before, we have a working configuration if and only if the maximum weight placed is always on the left side, so the number of ways to assign that is 
.
,
is uniquely determined by this recursion, so we show that this solution indeed works. Note that 
,
.
.
,
as desired.
for 
and thus 
placed to put it. Therefore, 
Since 
we can easily obtain that 
which is the answer.
in general. Now notice that the 
ways already. Now we could have placed 
weights is 
.
,
because it must be placed in the left pan. From the scenario of 
,
so we only need to choose where 
.
.
