Q.Prove that $^{35}C_5 + \displaystyle\sum_{r=0}^{4} {}^{(39-r)}C_4 = {}^{40}C_5$.

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: $n=\dfrac{3\pm\sqrt{9+352}}{2}=\dfrac{3\pm\sqrt{361}}{2}=\dfrac{3\pm19}{2}$.

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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.

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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 $x=1$:

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

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.

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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:

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

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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 $r=4$:

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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 $\binom{15}{2}=\dfrac{15\times14}{2}=105$.

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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:

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