Geometrical problems Questions in English

(B) An $8$-oar boat requires $4$ men on the stroke side and $4$ men on the bow side.
Given $12$ men,$3$ can row only on the stroke side,and $9$ can row on either side.
To form the crew,we need $4$ men for the stroke side and $4$ for the bow side.
Case $1$: Select $4$ men for the stroke side. We must include some of the $3$ stroke-only men.
Total ways to select $4$ men for the stroke side from $12$ is ${ }^{9}C_{4}$ (choosing from the $9$ flexible men) plus cases involving the $3$ stroke-only men.
Actually,the standard approach is: Select $4$ men for the stroke side from the $3$ stroke-only and $9$ others,then $4$ for the bow side from the remaining $8$.
Total ways $= { }^{9}C_{4} \times { }^{8}C_{4} \times 4! \times 4!$.
Thus,option $(B)$ is correct.

(C) The total number of ways to select $3$ points out of $10$ is given by $^{10}C_3$.
$^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
Since $6$ points are collinear,they do not form a triangle. The number of ways to select $3$ points from these $6$ collinear points is $^{6}C_3$.
$^{6}C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
Therefore,the total number of triangles $N = 120 - 20 = 100$.

(A) triangle is formed by selecting $3$ vertices out of $n$ vertices. The number of ways to choose $3$ vertices is given by ${}^nC_3$.
Given $T_{n+1} - T_n = 10$.
Using the identity ${}^{n+1}C_r - {}^nC_r = {}^nC_{r-1}$,we have:
${}^{n+1}C_3 - {}^nC_3 = {}^nC_2 = 10$.
Expanding the combination formula:
$\frac{n(n-1)}{2} = 10$
$n^2 - n = 20$
$n^2 - n - 20 = 0$
$(n-5)(n+4) = 0$
Since $n$ must be a positive integer,$n = 5$.

(D) The total number of distinct triangles that can be formed using $15$ points is given by $^{15}C_3 = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455$.
We need to exclude the cases where $i+j+k = 15$,where $1 \leq i < j < k \leq 15$.
The possible sets of $(i, j, k)$ such that $i+j+k = 15$ are:
$(1, 2, 12), (1, 3, 11), (1, 4, 10), (1, 5, 9), (1, 6, 8), (2, 3, 10), (2, 4, 9), (2, 5, 8), (2, 6, 7), (3, 4, 8), (3, 5, 7), (4, 5, 6)$.
Counting these,we find there are $12$ such sets.
Therefore,the number of required triangles is $455 - 12 = 443$.

(C) The total number of ways to select $2$ lines out of $37$ is $^{37}C_2$.
Since $13$ lines are concurrent at point $A$,they do not produce $^{13}C_2$ distinct intersection points; instead,they all intersect at $1$ point. Thus,we subtract $^{13}C_2$ and add $1$.
Similarly,since $11$ lines are concurrent at point $B$,they do not produce $^{11}C_2$ distinct intersection points; they all intersect at $1$ point. Thus,we subtract $^{11}C_2$ and add $1$.
Therefore,the total number of points of intersection is $^{37}C_2 - ^{13}C_2 + 1 - ^{11}C_2 + 1 = ^{37}C_2 - ^{13}C_2 - ^{11}C_2 + 2$.

(D) parallelogram is formed by selecting $2$ lines from the first set of $10$ parallel lines and $2$ lines from the second set of $m$ parallel lines.
The number of ways to choose $2$ lines from $10$ is given by ${}^{10}C_2 = \frac{10 \times 9}{2} = 45$.
The number of ways to choose $2$ lines from $m$ is given by ${}^{m}C_2 = \frac{m(m-1)}{2}$.
The total number of parallelograms is the product of these two combinations:
${}^{10}C_2 \times {}^{m}C_2 = 675$
$45 \times \frac{m(m-1)}{2} = 675$
$\frac{m(m-1)}{2} = \frac{675}{45} = 15$
$m(m-1) = 30$
$m^2 - m - 30 = 0$
$(m - 6)(m + 5) = 0$
Since $m$ must be positive,$m = 6$.

(C) Total points available excluding $A, B,$ and $C$ are $10 + 8 = 18$.
To form a triangle,we need to select $3$ points that are not collinear.
The total number of ways to select $3$ points from $18$ is $^{18}C_3$.
Points on line $AB$ are collinear,so $^{10}C_3$ combinations do not form a triangle.
Points on line $AC$ are collinear,so $^{8}C_3$ combinations do not form a triangle.
Thus,the number of triangles is $^{18}C_3 - ^{10}C_3 - ^{8}C_3$.
$^{18}C_3 = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816$.
$^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
$^{8}C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
Number of triangles $= 816 - 120 - 56 = 640$.

(A) Let the number of sides of the polygon be $n$. The number of diagonals in a polygon with $n$ sides is given by the formula $\frac{n(n-3)}{2}$.
Given that the number of diagonals is $54$,we have:
$\frac{n(n-3)}{2} = 54$
$n(n-3) = 108$
$n^2 - 3n - 108 = 0$
$(n-12)(n+9) = 0$
Since $n$ must be a positive integer,we have $n = 12$.
Thus,the polygon has $12$ sides.

(B) The number of diagonals in a polygon with $n$ sides is given by the formula $\frac{n(n-3)}{2}$.
Given that the number of diagonals is $170$,we have:
$\frac{n(n-3)}{2} = 170$
$n(n-3) = 340$
$n^2 - 3n - 340 = 0$
Solving the quadratic equation using the quadratic formula or factoring:
$(n - 20)(n + 17) = 0$
Since $n$ must be positive,$n = 20$.
The measure of each interior angle of a regular polygon with $n$ sides is given by $\frac{(n-2) \pi}{n}$.
For $n = 20$,the interior angle is $\frac{(20-2) \pi}{20} = \frac{18 \pi}{20} = \frac{9 \pi}{10}$.
Thus,the correct option is $B$.

(C) Let there be $n$ points on the plane. The number of line segments formed by connecting $n$ points pairwise is given by the combination formula ${}^nC_2$.
Given that the total number of line segments is $15$,we have:
${}^nC_2 = 15$
$\frac{n(n - 1)}{2} = 15$
$n(n - 1) = 30$
$n^2 - n - 30 = 0$
$(n - 6)(n + 5) = 0$
Since the number of points $n$ must be positive,we have $n = 6$.

(C) polygon is formed by selecting $n$ points out of $10$ points,where $n \ge 3$.
The number of ways to select $n$ points from $10$ is given by the combination formula $\binom{10}{n}$.
Since any selection of $n$ points $(n \ge 3)$ on a circle forms exactly one unique convex polygon,the total number of such polygons is the sum of combinations for $n = 3, 4, \dots, 10$.
Total polygons = $\binom{10}{3} + \binom{10}{4} + \binom{10}{5} + \binom{10}{6} + \binom{10}{7} + \binom{10}{8} + \binom{10}{9} + \binom{10}{10}$.
We know that $\sum_{n=0}^{10} \binom{10}{n} = 2^{10} = 1024$.
Therefore,$\sum_{n=3}^{10} \binom{10}{n} = 2^{10} - \binom{10}{0} - \binom{10}{1} - \binom{10}{2}$.
$= 1024 - 1 - 10 - \frac{10 \times 9}{2} = 1024 - 1 - 10 - 45 = 1024 - 56 = 968$.

(B) The set $X$ represents a grid of points $(a, b)$ where $a$ ranges from $0$ to $20$ ($21$ points) and $b$ ranges from $0$ to $15$ ($16$ points).
To form a rectangle with sides parallel to the axes,we need to choose $2$ distinct values for $a$ from the set $\{0, 1, 2, \dots, 20\}$ and $2$ distinct values for $b$ from the set $\{0, 1, 2, \dots, 15\}$.
The number of ways to choose $2$ values for $a$ is $\binom{21}{2} = \frac{21 \times 20}{2} = 210$.
The number of ways to choose $2$ values for $b$ is $\binom{16}{2} = \frac{16 \times 15}{2} = 120$.
The total number of such rectangles is $210 \times 120 = 25200$.

(C) There are $3$ parallel lines,each containing $4$ points. The total number of points is $3 \times 4 = 12$.
The total number of ways to select $3$ points from $12$ is given by $C(12, 3) = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
Triangles are not formed when the $3$ points are collinear. Collinear points occur when we select $3$ points from the same line.
Since there are $3$ lines each with $4$ points,the number of ways to choose $3$ collinear points is $3 \times C(4, 3) = 3 \times 4 = 12$.
Therefore,the number of triangles formed is $220 - 12 = 208$.

(C) The number of triangles formed by choosing $3$ vertices from $n$ vertices of a regular polygon is given by $t_n = \binom{n}{3} = \frac{n(n-1)(n-2)}{6}$.
Given the condition $t_{n+1} = t_n + 28$,we substitute the formula:
$\binom{n+1}{3} - \binom{n}{3} = 28$.
Using the property $\binom{n+1}{3} - \binom{n}{3} = \binom{n}{2}$,we get:
$\binom{n}{2} = 28$.
$\frac{n(n-1)}{2} = 28$.
$n(n-1) = 56$.
$n^2 - n - 56 = 0$.
$(n-8)(n+7) = 0$.
Since $n$ must be positive,$n = 8$.

(B) The number of triangles that can be formed with the vertices of a polygon of $m$ sides is given by $T_m = {}^mC_3$.
Given that $T_{m+1} - T_m = 15$.
Substituting the formula,we get ${}^{m+1}C_3 - {}^mC_3 = 15$.
Using the identity ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$,we can write ${}^{m+1}C_3 = {}^mC_3 + {}^mC_2$.
Therefore,${}^mC_3 + {}^mC_2 - {}^mC_3 = 15$.
This simplifies to ${}^mC_2 = 15$.
Expanding the combination,$\frac{m(m-1)}{2} = 15$.
$m(m-1) = 30$.
$m^2 - m - 30 = 0$.
$(m-6)(m+5) = 0$.
Since $m$ must be a positive integer,$m = 6$.

(D) Total number of points is $3p$.
To form a triangle,we need to select $3$ points out of $3p$.
The total number of ways to select $3$ points is $^{3p}C_3$.
However,if $3$ points are collinear,they do not form a triangle.
There are $3$ lines,each containing $p$ points.
For each line,the number of ways to select $3$ collinear points is $^pC_3$.
Thus,the number of triangles is $^{3p}C_3 - 3 \cdot ^pC_3$.
$= \frac{3p(3p-1)(3p-2)}{6} - 3 \cdot \frac{p(p-1)(p-2)}{6}$
$= \frac{p}{2} [ (3p-1)(3p-2) - (p-1)(p-2) ]$
$= \frac{p}{2} [ (9p^2 - 9p + 2) - (p^2 - 3p + 2) ]$
$= \frac{p}{2} [ 8p^2 - 6p ]$
$= p^2(4p-3)$.

(B) The number of diagonals of a polygon with $n$ sides is given by the formula: $\frac{n(n-3)}{2}$.
Given that the number of diagonals is $104$,we have:
$\frac{n(n-3)}{2} = 104$
$n(n-3) = 208$
$n^2 - 3n - 208 = 0$
Solving this quadratic equation using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$n = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-208)}}{2(1)}$
$n = \frac{3 \pm \sqrt{9 + 832}}{2}$
$n = \frac{3 \pm \sqrt{841}}{2}$
$n = \frac{3 \pm 29}{2}$
Since $n$ must be positive,$n = \frac{3 + 29}{2} = \frac{32}{2} = 16$.
Thus,$n = 16$.

(C) The total number of ways to choose $3$ squares from $64$ is $\binom{64}{3} = \frac{64 \times 63 \times 62}{3 \times 2 \times 1} = 41664$.
On a chess board,the diagonals have lengths $1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1$ in both directions.
For a diagonal of length $k$,the number of ways to choose $3$ squares is $\binom{k}{3}$.
We only consider diagonals with length $k \ge 3$.
The lengths are $3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3$.
There are two sets of these diagonals (one for each direction).
Sum of $\binom{k}{3}$ for one direction: $\binom{3}{3} + \binom{4}{3} + \binom{5}{3} + \binom{6}{3} + \binom{7}{3} + \binom{8}{3} + \binom{7}{3} + \binom{6}{3} + \binom{5}{3} + \binom{4}{3} + \binom{3}{3} = 1 + 4 + 10 + 20 + 35 + 56 + 35 + 20 + 10 + 4 + 1 = 196$.
Since there are two directions,the total number of favorable outcomes is $196 \times 2 = 392$.
The probability is $\frac{392}{41664} = \frac{7}{744}$.

(A) cube has $8$ vertices. The total number of ways to choose $3$ vertices out of $8$ is given by $^8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
An equilateral triangle is formed by choosing $3$ vertices such that the distance between any two of them is equal to the length of the face diagonal of the cube.
Each face of the cube has $2$ diagonals,and there are $6$ faces,but each diagonal is shared by $2$ faces. The total number of face diagonals is $12$.
However,an equilateral triangle is formed by selecting $3$ vertices such that each pair is connected by a face diagonal. There are exactly $8$ such triangles in a cube (each corresponding to a set of $3$ vertices that are mutually equidistant).
Therefore,the number of favorable outcomes is $8$.
The probability is $\frac{8}{56} = \frac{1}{7}$.

(C) Given that $3$ out of $6$ vertices of a regular hexagon are chosen.
Total number of ways to choose $3$ vertices out of $6$ is given by $^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
In a regular hexagon,only two equilateral triangles can be formed by joining its vertices,which are $\triangle ACE$ and $\triangle BDF$.
Thus,the number of favorable outcomes is $2$.
Therefore,the probability is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{20} = \frac{1}{10}$.

(C) The total number of points $(x, y)$ such that $0 \leq x \leq 4$ and $0 \leq y \leq 4$ is $(4+1) \times (4+1) = 25$ points.
To form a triangle,we need to select $3$ non-collinear points from these $25$ points.
The total number of ways to select $3$ points from $25$ is $\binom{25}{3} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 2300$.
Now,we must subtract the number of sets of $3$ collinear points.
Points are collinear if they lie on the same row,column,or diagonal.
$1$. Rows: There are $5$ rows,each with $5$ points. Number of ways to choose $3$ points from a row is $5 \times \binom{5}{3} = 5 \times 10 = 50$.
$2$. Columns: There are $5$ columns,each with $5$ points. Number of ways to choose $3$ points from a column is $5 \times \binom{5}{3} = 5 \times 10 = 50$.
$3$. Diagonals:
- Main diagonals: $2$ diagonals of length $5$ $(\binom{5}{3} \times 2 = 20)$.
- Other diagonals of length $4$: $4$ diagonals $(\binom{4}{3} \times 4 = 16)$.
- Other diagonals of length $3$: $4$ diagonals $(\binom{3}{3} \times 4 = 4)$.
Total collinear sets = $50 + 50 + 20 + 16 + 4 = 140$.
Number of triangles = $2300 - 140 = 2160$.

(C) Total points = $10$. Points that are collinear = $4$. Points that are non-collinear = $10 - 4 = 6$.
To form a triangle with at least one vertex from the $4$ collinear points,we consider the following cases:
Case $1$: Select $1$ point from $4$ collinear points and $2$ points from $6$ non-collinear points.
Number of ways = $\binom{4}{1} \times \binom{6}{2} = 4 \times 15 = 60$.
Case $2$: Select $2$ points from $4$ collinear points and $1$ point from $6$ non-collinear points.
Number of ways = $\binom{4}{2} \times \binom{6}{1} = 6 \times 6 = 36$.
Note: We cannot select $3$ points from the $4$ collinear points because they are collinear and will not form a triangle.
Total number of triangles = $60 + 36 = 96$.

(B) The number of diagonals of a polygon with $n$ vertices is given by $\frac{n(n-1)}{2} - n = 35$.
Solving for $n$:
$n^2 - n - 2n = 70$
$n^2 - 3n - 70 = 0$
$(n - 10)(n + 7) = 0$
Since $n$ must be positive,$n = 10$.
To form a triangle with $AB$ as one of its sides,we must choose $A$,$B$,and one other vertex from the remaining $n - 2$ vertices.
Number of remaining vertices $= 10 - 2 = 8$.
Therefore,the number of such triangles is $8$.

(C) The total number of points in a plane is $n = 30$.
Since $8$ points are collinear,they do not form distinct lines when joined in pairs,except for the single line they all lie on.
The formula for the number of lines formed by $n$ points where $m$ points are collinear is given by:
$\text{Number of lines} = {}^{n}C_{2} - {}^{m}C_{2} + 1$
Substituting the values $n = 30$ and $m = 8$:
$\text{Number of lines} = {}^{30}C_{2} - {}^{8}C_{2} + 1$
$= \frac{30 \times 29}{2} - \frac{8 \times 7}{2} + 1$
$= 435 - 28 + 1$
$= 408$

(C) We know that the number of diagonals in a polygon of $n$ sides is given by the formula: $\frac{n(n-3)}{2}$.
Given that the number of diagonals is $170$,we have:
$\frac{n(n-3)}{2} = 170$
$n(n-3) = 340$
$n^2 - 3n - 340 = 0$
Solving the quadratic equation by factoring:
$n^2 - 20n + 17n - 340 = 0$
$n(n - 20) + 17(n - 20) = 0$
$(n - 20)(n + 17) = 0$
Since $n$ must be positive,$n = 20$.

(D) The total number of points is $8$ (excluding $P$). Including $P$,we have $9$ points in total.
To form a triangle,we need to select $3$ non-collinear points.
The total number of ways to select $3$ points from $9$ points is ${^9C_3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
However,points on the same line cannot form a triangle.
Points on $l_1$ (including $P$) are $4$ $(A_1, B_1, C_1, P)$. The number of ways to choose $3$ collinear points from these is ${^4C_3} = 4$.
Points on $l_2$ (including $P$) are $6$ $(A_2, B_2, C_2, D_2, E_2, P)$. The number of ways to choose $3$ collinear points from these is ${^6C_3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
Total triangles = (Total ways to choose $3$ points) - (Collinear sets on $l_1$) - (Collinear sets on $l_2$)
Total triangles = $84 - 4 - 20 = 60$.

(D) The total number of ways to select $3$ girls from $15$ girls seated around a round table is given by the combination formula ${}^{15}C_3$.
First,we calculate the total number of ways to select $3$ girls: ${}^{15}C_3 = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 5 \times 7 \times 13 = 455$.
Next,we find the number of ways to select $3$ girls such that they are all seated together. In a circular arrangement of $15$ girls,there are $15$ possible sets of $3$ consecutive girls.
So,the number of ways to select $3$ girls sitting together is $15$.
Therefore,the required number of ways where all three are not seated together is $455 - 15 = 440$.

(D) triangle is formed by selecting any $2$ lines out of the $15$ concurrent lines and the line $L$ acting as the third side.
Since the $15$ lines are concurrent at point $P$,any pair of these lines will intersect at $P$.
When these $2$ lines intersect the line $L$ at two distinct points,a triangle is formed with $L$ as one of its sides.
The number of ways to choose $2$ lines out of $15$ is given by the combination formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
Thus,the number of triangles is ${}^{15}C_{2} = \frac{15 \times 14}{2 \times 1} = 105$.

(B) Total points = $10$. Points that are collinear = $4$. Points that are non-collinear = $6$.
To form a triangle with at least one vertex from the $4$ collinear points,we consider two cases:
Case $1$: One vertex from the $4$ collinear points and two vertices from the $6$ non-collinear points.
Number of triangles = $^4C_1 \times ^6C_2 = 4 \times 15 = 60$.
Case $2$: Two vertices from the $4$ collinear points and one vertex from the $6$ non-collinear points.
Number of triangles = $^4C_2 \times ^6C_1 = 6 \times 6 = 36$.
Total number of triangles = $60 + 36 = 96$.

(C) The number of diagonals of a regular polygon having $n$ sides is given by $\frac{n(n-3)}{2}$.
$\therefore \frac{n(n-3)}{2} = 35$
$\Rightarrow n(n-3) = 70$
$\Rightarrow n^2 - 3n - 70 = 0$
$\Rightarrow n^2 - 10n + 7n - 70 = 0$
$\Rightarrow n(n - 10) + 7(n - 10) = 0$
$\Rightarrow (n - 10)(n + 7) = 0$
Since the number of sides $n$ must be positive,we have $n = 10$.

(D) The number of diagonals in a polygon with $n$ sides is given by the formula $\frac{n(n-3)}{2}$.
Given that the number of diagonals is $54$,we have:
$\frac{n(n-3)}{2} = 54$
$n(n-3) = 108$
$n^2 - 3n - 108 = 0$
Factoring the quadratic equation:
$n^2 - 12n + 9n - 108 = 0$
$n(n - 12) + 9(n - 12) = 0$
$(n + 9)(n - 12) = 0$
This gives $n = -9$ or $n = 12$.
Since the number of sides $n$ must be a positive integer,we have $n = 12$.

(B) Total number of ways to arrange $6$ students in a row $= 6! = 720$.
To find the number of favorable arrangements,consider $A$ and $B$ with one student between them as a single block. There are $4$ remaining students. We choose $1$ student out of $4$ to place between $A$ and $B$ in $^4C_1$ ways.
Now,treat the block $(A, \text{student}, B)$ as one unit. Along with the remaining $3$ students,we have $4$ units to arrange,which can be done in $4!$ ways.
Within the block,$A$ and $B$ can be arranged in $2!$ ways (i.e.,$A, \text{student}, B$ or $B, \text{student}, A$).
Number of favorable arrangements $= ^4C_1 \times 4! \times 2! = 4 \times 24 \times 2 = 192$.
Probability $= \frac{192}{720} = \frac{4}{15}$.

(B) The number of diagonals in a regular polygon of $n$ sides is given by the formula $\frac{n(n-3)}{2}$ or ${ }^{n} C_{2}-n$.
For a polygon with $n = 100$ sides:
Number of diagonals $= { }^{100} C_{2} - 100$
$= \frac{100 \times 99}{2} - 100$
$= 50 \times 99 - 100$
$= 4950 - 100$
$= 4850$

(C) The number of diagonals in a polygon with $n$ sides is given by the formula: $\frac{n(n-3)}{2} = 20$.
Multiplying by $2$,we get $n(n-3) = 40$.
$n^2 - 3n - 40 = 0$.
Factoring the quadratic equation: $(n-8)(n+5) = 0$.
Since the number of sides $n$ must be positive,we have $n = 8$.

(B) The number of diagonals in a polygon with $n$ sides is given by the formula $\frac{n(n-3)}{2}$.
Given that the number of diagonals is $44$,we have:
$\frac{n(n-3)}{2} = 44$
$n(n-3) = 88$
$n^2 - 3n - 88 = 0$
$(n - 11)(n + 8) = 0$
Since the number of sides $n$ must be positive,we have $n = 11$.

(C) To form a pentagon using the given points,we must select $5$ points such that no three points are collinear. Since the points are on the sides of a triangle,we can select points from the sides $AB$,$BC$,and $AC$.
Case $1$: $2$ points from $AB$,$2$ points from $BC$,and $1$ point from $AC$.
Number of ways = $\binom{4}{2} \times \binom{5}{2} \times \binom{4}{1} = 6 \times 10 \times 4 = 240$.
Case $2$: $2$ points from $AB$,$1$ point from $BC$,and $2$ points from $AC$.
Number of ways = $\binom{4}{2} \times \binom{5}{1} \times \binom{4}{2} = 6 \times 5 \times 6 = 180$.
Case $3$: $1$ point from $AB$,$2$ points from $BC$,and $2$ points from $AC$.
Number of ways = $\binom{4}{1} \times \binom{5}{2} \times \binom{4}{2} = 4 \times 10 \times 6 = 240$.
Total number of pentagons = $240 + 180 + 240 = 660$.

(C) The number of triangles $p_n$ formed by joining the vertices of an $n$-sided polygon is given by the combination formula $\binom{n}{3}$.
Given the relation $p_{n+1} - p_n = 66$,we substitute the formula: $\binom{n+1}{3} - \binom{n}{3} = 66$.
Using the property $\binom{n+1}{k} - \binom{n}{k} = \binom{n}{k-1}$,we get $\binom{n}{2} = 66$.
Expanding the combination: $\frac{n(n-1)}{2} = 66$,which simplifies to $n^2 - n = 132$.
Rearranging gives the quadratic equation $n^2 - n - 132 = 0$.
Factoring the equation: $(n - 12)(n + 11) = 0$.
Since $n$ must be positive,$n = 12$.
The prime factorization of $12$ is $2^2 \times 3^1$.
The distinct prime divisors of $12$ are $2$ and $3$.
The sum of these prime divisors is $2 + 3 = 5$.