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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
UC Berkeley Integration Bee 2025 Bracket Rounds
Silver08   15
N 4 hours ago by vanstraelen
Regular Round

Quarterfinals

Semifinals

3rd Place Match

Finals
15 replies
Silver08
May 9, 2025
vanstraelen
4 hours ago
ISI UGB 2025 P3
SomeonecoolLovesMaths   7
N 4 hours ago by MathsSolver007
Source: ISI UGB 2025 P3
Suppose $f : [0,1] \longrightarrow \mathbb{R}$ is differentiable with $f(0) = 0$. If $|f'(x) | \leq f(x)$ for all $x \in [0,1]$, then show that $f(x) = 0$ for all $x$.
7 replies
SomeonecoolLovesMaths
Today at 11:32 AM
MathsSolver007
4 hours ago
Minimum value
Martin.s   3
N 5 hours ago by Martin.s
What is the minimum value of
$$
\frac{|a + b + c + d| \left( |a - b| |b - c| |c - d| + |b - a| |c - a| |d - a| \right)}{|a - b| |b - c| |c - d| |d - a|}
$$over all triples $a, b, c, d$ of distinct real numbers such that
$a^2 + b^2 + c^2 + d^2 = 3(ab + bc + cd + da).$

3 replies
Martin.s
Oct 17, 2024
Martin.s
5 hours ago
Cost Question
bassamali01   9
N Today at 4:19 PM by Juno_34
Sorry, I have been struggling with this question so much. It is a simple derivative question as I think it is. Can I get some help on it?
9 replies
bassamali01
Dec 7, 2017
Juno_34
Today at 4:19 PM
No more topics!
Subset Ordered Pairs of {1, 2, ..., 10}
ahaanomegas   11
N Yesterday at 5:27 AM by cappucher
Source: Putnam 1990 A6
If $X$ is a finite set, let $X$ denote the number of elements in $X$. Call an ordered pair $(S,T)$ of subsets of $ \{ 1, 2, \cdots, n \} $ $ \emph {admissible} $ if $ s > |T| $ for each $ s \in S $, and $ t > |S| $ for each $ t \in T $. How many admissible ordered pairs of subsets $ \{ 1, 2, \cdots, 10 \} $ are there? Prove your answer.
11 replies
ahaanomegas
Jul 12, 2013
cappucher
Yesterday at 5:27 AM
Subset Ordered Pairs of {1, 2, ..., 10}
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G H BBookmark kLocked kLocked NReply
Source: Putnam 1990 A6
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ahaanomegas
6294 posts
#1 • 2 Y
Y by Adventure10, Mango247
If $X$ is a finite set, let $X$ denote the number of elements in $X$. Call an ordered pair $(S,T)$ of subsets of $ \{ 1, 2, \cdots, n \} $ $ \emph {admissible} $ if $ s > |T| $ for each $ s \in S $, and $ t > |S| $ for each $ t \in T $. How many admissible ordered pairs of subsets $ \{ 1, 2, \cdots, 10 \} $ are there? Prove your answer.
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JBL
16123 posts
#2 • 2 Y
Y by Adventure10, Mango247
Here are some past discussions of this problem:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=46819&hilit=Putnam+1990
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=515970&hilit=Putnam+1990
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IAmTheHazard
5001 posts
#3 • 2 Y
Y by natmath, centslordm
I suppose this is the official thread? It's not in the putnam collection, nor is the rest of the 1990 putnam...

The answer is $F_{2n+2}$ where $(F_n)$ denotes the Fibonacci sequence, with $F_1=F_2=1$ and $F_k=F_{k-1}+F_{k-2}$. For some $n$, by summing over $|A|=a$ and $|B|=b$, it is clear that there are
$$\sum_{a\geq 0}\sum_{b\geq 0}\binom{n-a}{b}\binom{n-b}{a}=a_n$$ways to choose $A$ and $B$. Thus consider the formal series
$$A(x)=\sum_{n\geq 0}a_nx^n=\sum_{n\geq 0}\sum{a\geq 0}\sum_{b\geq 0}\binom{n-b}{a}\binom{n-a}{b}x^n.$$Then, we have
\begin{align*}
    A(x)&=\sum_{a\geq 0}\sum_{b\geq 0}\sum_{n\geq a+b}\binom{n-b}{a}\binom{n-a}{b}x^n\\
    &=\sum_{a\geq 0}\sum_{b\geq 0}\sum_{m\geq 0}\binom{m+a}{a}\binom{m+b}{b}x^{m+a+b}\\
    &=\sum_{m\geq 0}x^m\left(\sum_{a\geq 0}\binom{m+a}{a}x^a\right)^2\\
    &=\sum_{m\geq 0}\frac{x^m}{(1-x)^{2m+2}}\\
    &=\frac{\frac{1}{(1-x)^2}}{1-\frac{x}{(1-x)^2}}=\frac{1}{(1-x)^2-x}\\
    &=\frac{1}{1-3x+x^2}=\frac{1}{x\sqrt{5}}\left(\frac{1}{1-x\alpha}-\frac{1}{1-x\beta}\right)
\end{align*}where $\alpha=\tfrac{3+\sqrt{5}}{2}$ and $\beta=\tfrac{3-\sqrt{5}}{2}$ are the roots of $x^2-3x+1=0$. By the geometric series formula it follows that $a_n=\tfrac{1}{\sqrt{5}}(\alpha^{n+1}-\beta^{n+1})$.
Finally, note that $\alpha=(\tfrac{1+\sqrt{5}}{2})^2$ and $\beta=(\tfrac{1-\sqrt{5}}{2})^2$, so we have
$$a_n=\frac{1}{\sqrt{5}}(\alpha^{n+1}-\beta^{n+1})=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{2n+2}-\left(\frac{1-\sqrt{5}}{2}\right)^{2n+2}\right)=F_{2n+2}$$as desired. $\blacksquare$
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HamstPan38825
8863 posts
#4
Y by
The number in question is precisely equal to $$\sum_{i \geq 0}\sum_{j \geq 0}{n-i \choose j}{n-j \choose i}.$$We sum this by parts according to $i+j = k$. Eaach of these parts is equal to $$ \sum_{i+j = k} {n-i \choose n-k}{n-j \choose n-k} = \sum_{p+q=2n-k} {p \choose n-k}{q \choose n-k}.$$Here $p=n-i$ and $q = n-j$. This is just the $2n-k$ coefficient in $$f(x) = \left(\sum_{p \geq 0} {p \choose n-k} x^p\right)^2 = \frac{x^{2n-2k}}{(1-x)^{2n-2k+2}}.$$Expanding out the series for this, the desired coefficient is ${2n-k+1 \choose k}$. Hence, the answer is $$\sum_{k=0}^{2n} {2n-k+1 \choose k} = \sum_{k=0}^{2n+1} {2n+1-k \choose k} = F_{2n+2}.$$
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Shreyasharma
682 posts
#5
Y by
If the cardinality of $A$ and $B$ are $i$ and $j$ respectively, we find we need to choose $i$ numbers from $(j + 1, \dots, n)$ numbers to fill set $A$ and $j$ numbers from $(i + 1, \dots, n)$ numbers to fill set $B$. Thus we have $\binom{n-i}{j}$ choices for $A$ and $\binom{n-j}{i}$ choices for $B$. Note we have the added restriction that $i + j \leq n$, else the binomial coefficients are undefined.
\begin{align*}
F(n) = \sum_{i + j \leq n} \binom{n-i}{j} \cdot \binom{n-j}{i}
\end{align*}Then we will use snake oil. The idea is to let $i + j + k = n$, and consider the generating function,
\begin{align*}
A(X) = \sum_{n \geq 0} F(n)X^n
\end{align*}Now by substituting $i +j + k = n$ we find,
\begin{align*}
A(X) = \sum_{k \geq 0} \sum_{i \geq 0} \sum_{j \geq 0} \binom{k+j}{j} \binom{k+i}{i} X^{i+j+k}
\end{align*}Now we find,
\begin{align*}
A(X) = \sum_{k \geq 0} \left( \sum_{i \geq 0} \binom{k+i}{i} X^i \right)^2 X^k
\end{align*}This then simplifies to,
\begin{align*}
A(X) = \sum_{k \geq 0} \frac{X^k}{(1-X)^{2k+2}}
\end{align*}Which in turn becomes,
\begin{align*}
A(X) = \frac{\frac{1}{(1-X)^2}}{1-\frac{X}{(1-X)^2}} &= \frac{1}{(1-X)^2 - X}\\
&= \frac{1}{X^2 - 3X + 1}\\
&= \frac{1}{X\sqrt{5}} \cdot  \left( \frac{1}{(1-\alpha X)} - \frac{1}{(1-\beta X)} \right)
\end{align*}where we take $\alpha = \frac{3+\sqrt{5}}{2}$ and $\beta = \frac{3 - \sqrt{5}}{2}$. Then we can use geometric series to find,
\begin{align*}
A(X) &= \frac{1}{X\sqrt{5}} \cdot \left( \left(1 + \alpha X + \alpha^2 X^2 + \dots \right) - \left(1 + \beta X + \beta^2 X^2 + \dots \right) \right)\\
&= \frac{1}{\sqrt{5}} \cdot \left( (\alpha - \beta) + (\alpha^2 - \beta^2)X + (\alpha^3 - \beta^3)X^2 \dots \right)
\end{align*}Now we want to find the coefficient of $X^n$. Thus the coefficient of $X^n$ is,
\begin{align*}
F(n) = \frac{1}{\sqrt{5}} \cdot (\alpha^{n+1} - \beta^{n})
\end{align*}
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shendrew7
796 posts
#6 • 1 Y
Y by Sagnik123Biswas
Let $|A| = a$ and $|B| = b$. Define the generating function $A(x) = \sum_{n \ge 0} a_nx^n$, where $a_n$ is the number of admissible pairs. This function can also be expressed as
\begin{align*}
A(x) &= \sum_{n \ge 0} \sum_{a \ge 0} \sum_{b \ge 0} \binom{n-a}{b} \binom{n-b}{a} x^n \\
&= \sum_{a \ge 0} \sum_{b \ge 0} \binom{n-a}{b} \left(\frac{x}{1-x}\right)^a x^b \cdot \frac{1}{1-x} \\
&= \frac{(1+x)^n}{1-x} \cdot \frac{1}{1 - \frac{x}{1-x^2}} \\
&= \frac{(1+x)^{a+1}}{1-x-x^2} \\
&= \sum_{n \ge 0} \sum_{k \ge 0} F_{k+1} \binom{n+1}{k+1} x^n \\
&= \frac{1}{(1-x)^2} \cdot \sum_{k \ge 0} F_{k+1} \left(\frac{x}{1-x}\right)^k \\
&= \frac{1}{(1-x)^2} \cdot \frac{1}{1 - \frac{x}{1-x} - \left(\frac{x}{1-x}\right)^2} \\
&= \frac{1}{1-3x+x^2} \\
&= \sum_{n \ge 0} \boxed{F_{2n+2}} \cdot x^n. \quad \blacksquare
\end{align*}
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Nguyen
54 posts
#7
Y by
Denote $a=|S|$, $b=|T|$.

Since $s>b$ for each $s$$\in$$S$, $S$ is a subset of $\{b+1, ..., n\}$ ($\emptyset$ if $b=n$).
Answer the following questions:
$b+1$$\in$$S?, ..., n$$\in$$S?$ These are $n-b$ questions in total.
Write a 2 for each "yes" and a 1 for each "no".
Then you have written $a$ 2's and $n-a-b$ 1's.

Similarly, answer the following questions:
$a+1$$\in$$T?, ..., n$$\in$$T?$ These are $n-a$ questions in total.
Write a 2 for each "yes" and a 1 for each "no".
Then you have written $b$ 2's and $n-a-b$ 1's.

You have two strings of 1's and 2's, concatenate the first string, a 1 as separator, and the second string.
This gives a total of $a+b$ 2's and $2(n-a-b)+1$ 1's, which sum to $2n+1$.
Therefore the amount is $F_{2n+2}$.

Conversely, any 1,2-string which sum to $2n+1$ must have an odd number of 1's, and the middle 1 is the separator.
Therefore the value of $a$ and $b$ can be determined, and the two sets can be uniquely recovered.
This post has been edited 1 time. Last edited by Nguyen, Apr 7, 2024, 3:37 AM
Reason: tex
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OronSH
1746 posts
#8 • 1 Y
Y by megarnie
The sum we want is \[\sum_{a\ge 0}\sum_{b\ge 0}\binom{n-b}a\binom{n-a}b=\sum_{a\ge 0}\sum_{b\ge 0}\binom{n-b}{n-a-b}\binom{n-a}{n-a-b}=\sum_{k\ge 0}\sum_{a+b=k}\binom{n-b}{n-k}\binom{n-a}{n-k}.\]Now we claim \[\sum_{a+b=k}\binom am\binom bn=\binom{k+1}{m+n+1}.\]This follows from \[\sum_{k\ge 0}\sum_{a+b=k}\binom am\binom bnx^k=\left(\sum_{a\ge 0}\binom amx^a\right)\left(\sum_{b\ge 0}\binom bnx^b\right)=\frac{x^m}{(1-x)^{m+1}}\cdot\frac{x^n}{(1-x)^{n+1}}=\frac{x^{m+n}}{(1-x)^{m+n+2}}=\sum_{c\ge 0}\binom{c+1}{m+n+1}x^c.\]Thus we want \[\sum_{k\ge 0}\binom{2n-k+1}{2n-2k+1}=\sum_{k\ge 0}\binom{2n-k+1}k.\]Now we claim \[\sum_{k\ge 0}\binom{a-k}k=F_{a+1}.\]This is because \[\sum_{a\ge 0}\sum_{k\ge 0}\binom{a-k}kx^a=\sum_{k\ge 0}\sum_{a\ge 0}\binom{a-k}kx^a=\sum_{k\ge 0}\frac{x^{2k}}{(1-x)^{k+1}}=\frac1{1-x}\left(\frac1{1-\frac{x^2}{1-x}}\right)=\frac1{1-x-x^2}=\sum_{a\ge 0}F_{a+1}x^a.\]Thus the answer is $F_{2n+2}$.
This post has been edited 2 times. Last edited by OronSH, Aug 22, 2024, 4:34 PM
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kiyoras_2001
678 posts
#9
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A related problem (not too closely, but anyway related).
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lpieleanu
3001 posts
#10
Y by
Solution
This post has been edited 1 time. Last edited by lpieleanu, Dec 16, 2024, 3:02 PM
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Ilikeminecraft
630 posts
#11
Y by
It can be seen that the answer is $F_{2n + 2}$ by using partial fractions and then comparing with the formula of Fibonacci numbers.
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cappucher
96 posts
#12
Y by
We claim the answer is $\boxed{F_{2n + 2}}$, where $F_{n}$ denotes the $n$th Fibonacci number.

Lemma 1: The final answer can be represented with the sum

\[S_{n} = \sum_{A \geq 0}\sum_{B \geq 0} \binom{n - B}{A} \binom{n - A}{B}\]
proof: We will prove this by fixing $|A|$ and $|B|$, deriving a closed form for this given $|A|$ and $|B|$, and sum over all possible pairs of $|A|$ and $|B|$. Note that for valid elements of set $A$, we must choose $A$ elements that are NOT in the range $1$ to $B$; thus, there are $\binom{n - B}{A}$ combinations. The same logic holds for choosing elements from set $B$. Since these selections are independent, we multiply these to obtain the result

\[\binom{n - B}{A}\binom{n - A}{B}\]
Summing over all possibilities yields our desired result.

To compute $S_{n}$, we use generating functions. Define $F(x)$ to be

\[F(x) = \sum_{n \geq 0} S_{n} x^n \]
Equivalently,

\[F(x) = \sum_{n \geq 0} \sum_{A \geq 0}\sum_{B \geq 0} \binom{n - B}{A} \binom{n - A}{B} x^n\]
We rearrange the summations and shift the indexing of the sum by $A + B$ to obtain

\[F(x) = \sum_{A \geq 0} \sum_{B \geq 0}\sum_{n \geq 0} \binom{n + A}{A} \binom{n + B}{B} x^{A + B + n}\]
The rest of the problem is a matter of algebra and identifying various well known generating functions:

\[F(x) = \sum_{n \geq 0} x^{n} \sum_{B \geq 0} \binom{n + B}{B} x^{B} \sum_{A \geq 0} \binom{n + A}{A} x^{A}\]\[F(x) = \sum_{n \geq 0} x^{n} \left( \sum_{A \geq 0} \binom{n + A}{A} x^{A} \right)^2 \]\[F(x) = \sum_{n \geq 0} x^{n} \left( \frac{1}{(1 - x)^{n + 1}} \right)^2 \]\[F(x) = \frac{1}{(1 - x)^2} \sum_{n \geq 0} \left( \frac{x}{(1 - x)^{2}} \right)^n \]
We now use the formula for the geometric series:

\[F(x) = \frac{1}{(1 - x)^2} \cdot \frac{1}{1 - \frac{x}{(1 - x)^2}} \]\[F(x) =  \frac{1}{(1 - x)^2 -x} = \frac{1}{x^2 - 3x + 1} \]
Define $\alpha = \frac{3 + \sqrt{5}}{2} = \left(\frac{1 + \sqrt{5}}{2}\right)^2$ and $\beta = \frac{3 - \sqrt{5}}{2} = \left(\frac{1 - \sqrt{5}}{2}\right)^2$; these are the roots of the quadratic in the denominator of $F(x)$. Then $F(x)$ can be written as

\[F(x) = \frac{1}{\sqrt{5}} \left(\frac{1}{x - \alpha} - \frac{1}{x - \beta}\right)\]\[F(x) = \sum_{n \geq 0} \frac{1}{x\sqrt{5}} \left(\frac{1}{1 - \frac{\alpha}{x}} - \frac{1}{1 - \frac{\beta}{x}}\right) x^n\]
However, since the polynomial's coefficients are symmetric, we can replace $\frac{1}{x}$ with $x$:

\[F(x) = \sum_{n \geq 0} \frac{1}{x\sqrt{5}} \left(\frac{1}{1 - x \alpha} - \frac{1}{1 - x \beta}\right) x^n\]\[F(x) = \sum_{n \geq 0} \frac{1}{\sqrt{5}} \left( \alpha^{n + 1} + \beta^{n + 1} \right) x^n\]\[F(x) = \sum_{n \geq 0} \frac{1}{\sqrt{5}} \left( \left(\left(\frac{1 + \sqrt{5}}{2}\right)^2\right)^{n + 1} + \left(\left(\frac{1 - \sqrt{5}}{2}\right)^2\right)^{n + 1} \right) x^n\]\[F(x) = \sum_{n \geq 0} \frac{1}{\sqrt{5}} \left( \left(\frac{1 + \sqrt{5}}{2}\right)^{2n + 2} + \left(\frac{1 - \sqrt{5}}{2}\right)^{2n + 2} \right) x^n\]
\[F(x) = \sum_{n \geq 0} F_{2n + 2} x^n\]
Corresponding coefficients of the final generating function to the original function proves that $S_n = F_{2n + 2}$, as desired.
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