Q.A cricket team of $11$ players is to be chosen from $15$ players, of whom $5$ are bowlers. In how many ways can the team be chosen so that it includes exactly $3$ bowlers?

Solving $\dfrac{n(n-3)}{2}=44$ gives $n=11$.

The number of diagonals of an $n$-sided polygon is $\dfrac{n(n-3)}{2}$.

Set this equal to 44: $\dfrac{n(n-3)}{2}=44 \Rightarrow n(n-3)=88 \Rightarrow n^2-3n-88=0$.

Solving by the quadratic formula: $n=\dfrac{3\pm\sqrt{9+352}}{2}=\dfrac{3\pm\sqrt{361}}{2}=\dfrac{3\pm19}{2}$.

Taking the positive root (since $n>0$): $n=\dfrac{22}{2}=11$.

Check: $\dfrac{11\times8}{2}=44$. ✓

Matching n keys to n locks one lock at a time, the worst-case total number of trials needed is $\dfrac{n(n+1)}{2}$.

Work on the locks one at a time, always trying the remaining untried keys on the current lock.

For the first lock, there are n candidate keys. In the worst case you try n-1 wrong keys first, and even the final (guaranteed correct) key still needs to be physically inserted and turned to open the lock, so opening the first lock can take up to n trials.

For the second lock, only n-1 keys remain (the correct key for lock 1 is now used up), so it takes up to n-1 trials.

Continuing, the k-th lock takes up to n-k+1 trials, until the very last lock, which has only 1 key left and needs exactly 1 trial.

Adding the worst case for every lock:

$n+(n-1)+(n-2)+\dots+2+1=\dfrac{n(n+1)}{2}.$

The sum of all the binomial coefficients of order n equals $2^n$.

By the binomial theorem, $(1+x)^n={}^nC_0+{}^nC_1x+{}^nC_2x^2+\dots+{}^nC_nx^n$.

Putting $x=1$:

$(1+1)^n={}^nC_0+{}^nC_1+{}^nC_2+\dots+{}^nC_n$

$2^n={}^nC_0+{}^nC_1+\dots+{}^nC_n.$

Total pairs give lines, but the 4 collinear points would be over-counted as many lines when they actually lie on just one line, so we subtract the excess.

If no 3 points were collinear, every pair of the 8 points would give a distinct line: $^8C_2 = 28$ lines.

But 4 of the points are collinear. Among just those 4 points, $^4C_2 = 6$ pairs were counted as 6 different lines above, when in fact all 4 points lie on the SAME single line.

So we must subtract the 6 over-counted lines and add back the 1 actual line they form:

$\text{Lines} = {}^8C_2 - {}^4C_2 + 1 = 28 - 6 + 1 = 23$

Solving $\dfrac{n(n-3)}{2}=44$ gives $n=11$.

A polygon with $n$ sides has $n$ vertices, and the number of diagonals is (total line segments between vertices) minus (the $n$ sides themselves):

$\text{diagonals} = {}^nC_2 - n = \frac{n(n-1)}{2}-n = \frac{n(n-3)}{2}$

Set this equal to 44:

$\frac{n(n-3)}{2}=44 \implies n(n-3)=88 \implies n^2-3n-88=0$

Using the quadratic formula: $n = \dfrac{3\pm\sqrt{9+352}}{2} = \dfrac{3\pm 19}{2}$, giving $n=11$ or $n=-8$ (rejected, sides can't be negative).

Pascal's rule turns the left side into $^nC_4$, and comparing $^nC_4>{}^nC_3$ using the ratio of consecutive combinations gives $n>7$.

By Pascal's identity, $^{n-1}C_3+{}^{n-1}C_4 = {}^nC_4$. So the given inequality becomes:

${}^nC_4 > {}^nC_3$

The ratio of consecutive binomial coefficients is $\dfrac{{}^nC_r}{{}^nC_{r-1}} = \dfrac{n-r+1}{r}$. For $r=4$:

$\frac{{}^nC_4}{{}^nC_3} = \frac{n-3}{4}$

For $^nC_4>{}^nC_3$, we need this ratio to exceed 1:

$\frac{n-3}{4}>1 \implies n-3>4 \implies n>7$

Number of lines $=\binom{15}{2}-\binom{5}{2}+1=96$.

Any 2 points determine a unique line, so if all 15 points were in 'general position' (no 3 collinear), the number of lines would be $\binom{15}{2}=\dfrac{15\times14}{2}=105$.

But 5 of the points are collinear, meaning every pair chosen from those 5 actually lies on the SAME single line, not 5 distinct lines' worth of pairs. Ordinarily $\binom{5}{2}=10$ pairs from those 5 points would be counted as 10 different lines, but they're really just 1 line. So we subtract the 10 over-counted lines and add back the 1 genuine line they form: $105-10+1=96$.

Using the diagonal-count formula $\frac{n(n-3)}{2}$ with $n=8$ gives 20 diagonals.

An $n$-gon has $n$ vertices. The total number of line segments joining any two vertices is ${}^nC_2 = \dfrac{n(n-1)}{2}$; subtracting the $n$ sides (which are not diagonals) gives the diagonal count:

$\text{Diagonals} = \dfrac{n(n-1)}{2} - n = \dfrac{n(n-1)-2n}{2} = \dfrac{n(n-3)}{2}$

For an octagon, $n=8$: $\dfrac{8\times5}{2} = \dfrac{40}{2} = 20$.