Problematic Problems

Paramjit, the following answers the last one.
Consider the subsets that have the element "$1$" in them. Clearly,there would be $\binom{n-1}{r-1}$ such subsets created by choosing remaining $r-1$ elements from the remaining $n-1$ elements.Obviously, in all such subsets "$1$" will be the smallest element.
Now for the subsets with 2. There would again be a total of$\binom{n-1}{r-1}$ subsets out of which all do not satisfy the condition that "$2$ is the smallest element. We want only those subsets which have '$2$" as he smallest element i.e., the subsets should not have the element "$1$" in them. There would be $\binom{n-2}{r-1}$ such subsets obtained by choosing $r-1$ elements of $n-2$ elements ( i.e., excluding $1$ and $2$ ).

For the subsets with "$3$" as the smallest element we would have to make subsets of $r-1$ elements out of $n-3$ elements (i.e., excluding $1, 2$ and $3$ ) giving us a $\binom{n-3}{r-1}$ subsets.
Similarly for "$4$" there would be $\binom{n-4}{r-1}$ subsets.
This idea can now be easily extrapolated for any $k^{th}$ element for $1 \leq k \leq (n-r+1)$.

Now, for the Arithmetic mean.........
Total number of elements = $\binom{n}{r}$
Therefore, $F(n,r) = \frac{ \sum_{k=1}^{n-r+1} k\binom{n-k}{r-1}}{\binom{n}{r}}$

The sum ( let it be $S$ ) can be more explicitly written as,
$S = \binom{n-1}{r-1}+ 2\binom{n-2}{r-1} + 3\binom{n-3}{r-1} + \dots + (n-r+1)\binom{r-1}{r-1}$

$S = \Bigg( \binom{n-1}{r-1} + \binom{n-2}{r-1} + \binom{n-3}{r-1} + \dots + \binom{r-1}{r-1} \Bigg)$

$+ \Bigg( \binom{n-2}{r-1} + \binom{n-3}{r-1} + \binom{n-4}{r-1} + \binom{n-5}{r-1} + \dots + \binom{r-1}{r-1} \Bigg)$

$+ \dots \dots + \Bigg( \binom{r-2}{r-1} + \binom{r-1}{r-1} \Bigg) + \Bigg(\binom{r-1}{r-1}\Bigg)$

Using the standard sum $\binom{n}{r} + \binom{n-1}{r} + \binom{n-2}{r} +\binom{n-3}{r} + \dots + \binom{r}{r} = \binom{n+1}{r+1}$, we can evaluate the above expression.

$\Rightarrow S = \bigg(\binom{n}{r}\bigg) + \bigg(\binom{n-1}{r}\bigg) + \bigg(\binom{n-2}{r}\bigg) + \bigg(\binom{n-3}{r}\bigg) + \dots + \bigg(\binom{r}{r}\bigg)$

$\Rightarrow S = \binom{n+1}{r+1}$

Therefore, $F(n,r) = \frac{\binom{n+1}{r+1}}{\binom{n}{r}} \Rightarrow F(n,r) = \frac{n+1}{r+1}$

Note: I just used the property of the sum of combinatorial coefficients directly. Just in case you its proof as well, I can add that too. Also in the proof if we slightly change our method we can directly obtain $S$.

For the fifth one, @Josh Rowley has provided an excellent solution in this note.

I guess it can be done in 22 ways. considering the rotation of 11 chairs as different cases. Here is my logic highest number is 11. So to his left and right only 9 and 10 can occupy. In similar logic 10 has left one side vacant which can be occupied by only 8. Going on further ,6,4,2,1 will be adjacent. Hence possible solution is 1,2,4,6,8,10,11,9,7,3,5 or 1,3,5,7,9,11,10,8,6,4,,2.
For all these 2 cases we can have 11 different chair positions each. Hence the total possible way must be 22 considering that each chair position is different.If rotation doesnt ,matters its 2

54

The first one (quite a cool result):

I assume that being in a circle the seats themselves are not numbered, so if you can rotate one seating so that it becomes another, these 2 will in fact be considered the same.

Firstly, consider the question with only 3 people. We clearly only have 2 options (in clockwise order): $(1,2,3)$ $(1,3,2)$

Consider what happens now when we had a 4th person. There are only 2 numbers in the set ${ 1,2,3 }$ which are either 1 or 2 away from 4, namely 2 and 3. More generally, if we have a valid arrangement with $n$ cricketers and then an $n+1$th cricketer appears, he can only sit in one place - between cricketer $n$ and cricketer $n-1$. Therefore we have the trivial recursion that $f(n+1) = f(n)$ when $n \geq 3$ Therefore the number of seatings for 11 cricketers is $f(11) = f(3) = \fbox{2}$

My proof for question 3 is kind of ugly, so I'm hoping someone can improve it. First, we may assume $a \ge 0$ since otherwise we could look at $-f(x)$ instead of $f(x)$. We can also assume that $b \ge 0$ since otherwise we could look at $f(-x)$ instead of $f(x)$ (note that this won't change $a$).

In this case, it's clear that the maximum value of $f(x)$ on $[-1,1]$ will occur at one of the endpoints, so all we have to show is that the minimum is greater than $-1.5$. It's enough to prove that this is true if $m = {\rm min}(f(-1), f(0), f(1)) = -1$, because otherwise $f(x) - (m+1)$ will still satisfy the conditions of the problem plus our extra condition, and will have a smaller minimum than $f(x)$, since $m+1$ would be positive. Also, we can assume that $b \le 2a$, because the minimum value for $f(x)$ is either at one of the endpoints, in which case there is nothing to prove, or at its one local minimum $x = -b/2a$. So the only meaningful case is when that minimum $x$-value is greater than $-1$ (it's negative by the above assumptions on $a$ and $b$). In this case the minimum value is $-\frac{b^2}{4a}-c$.

The conditions of the problem imply the three inequalities $-1 \le a-b+c \le 1$, $-1 \le c \le 1$, and $-1 \le a+b+c \le 1$. We will use these repeatedly.

There are three cases. Case 1: $f(1) = -1$. Then $a+b+c = -1$ and $a-b+c \ge -1$ implies that $b \le 0$, but we've assumed $b \ge 0$, so $b = 0$. Now $a+c = -1$, $a \ge 0$, and $c \ge -1$ imply equality, so $f(x) = -1$, with a minimum greater than $-1.5$.

Case 2: $f(0) = -1$, so $c = -1$. So the minimum value is $-\frac{b^2}{4a} - 1$. Now we use $a-b-1 \ge -1$ and $a+b-1 \le 1$ to get $a \ge b$ and $a+b \le 2$. Note that this implies $b \le 1$. So $\frac{b}{a} \le 1$ and $\frac{b}4 \le \frac14$, so $\frac{b^2}{4a} \le \frac14$, so the minimum is $\ge -\frac54$. (Equality holds when $a = b = -c = 1$.)

Case 3: $f(-1) = -1$. So $a-b+c = -1$. We use the two inequalities $c \ge -1$ and $a+b+c \le 1$ against the equality to get $a \le b$ and $b \le 1$. Note that the minimum here is $-\frac{b^2}{4a} - (b-1-a) = -\frac{(2a-b)^2}{4a} - 1$. Now $2a-b \le a \le 1$, so $(2a-b)^2 \le a \cdot 1$, so $\frac{(2a^-b)^2}{4a} \le \frac14$, so once again the minimum is at least $-\frac54$. (Equality holds when $a = b = -c = 1$ as above.)

If I got all this right, I'm pretty sure I proved that $f(x) \in [-1.25,1.25]$, which is the strongest possible (since if $f(x) = x^2+x-1$, $f(-0.5) = -1.25$).

For second Question

Lets say base of triangle is B and height is H

Largest possible rectangle will be square

Its Side = B*H/(B+H)