Geometrical problems Questions in English

(C) To be divisible by $5$,the unit place must be occupied by the digit $5$.
Since the number must lie between $3000$ and $4000$,the thousandth place must be occupied by the digit $3$.
We have $4$ remaining digits ${1, 2, 4, 6}$ to fill the remaining $2$ positions (hundreds and tens).
The number of ways to arrange these $2$ digits from the $4$ available is given by $^4P_2$.
Thus,the total number of such numbers is $^4P_2 = 4 \times 3 = 12$.

(C) First,arrange the $5$ boys in a row. They can be arranged in $5!$ ways.
After arranging the boys,we create gaps between them to ensure no two girls are together.
The arrangement of $5$ boys creates $6$ possible gaps (including the ends): $\_ B_1 \_ B_2 \_ B_3 \_ B_4 \_ B_5 \_$.
We have $3$ girls to be placed in these $6$ gaps.
The number of ways to choose and arrange $3$ girls in $6$ gaps is given by $^6P_3$.
Therefore,the total number of ways is $^6P_3 \times 5!$.

(C) First,arrange the $5$ boys in a row,which can be done in $5!$ ways.
This creates $6$ possible gaps (including the ends) where the $5$ girls can be placed to ensure no two girls are together: $\_ B_1 \_ B_2 \_ B_3 \_ B_4 \_ B_5 \_$.
The number of ways to arrange $5$ girls in these $6$ gaps is given by $^6P_5 = \frac{6!}{(6-5)!} = 6!$.
Therefore,the total number of arrangements is $5! \times 6!$.

(A) To ensure no two Hindi books are together,we first arrange the $21$ English books in a row.
This creates $22$ possible gaps (including the ends) where the $19$ Hindi books can be placed.
The number of ways to choose $19$ gaps out of $22$ is given by the combination formula $^{22}C_{19}$.
$^{22}C_{19} = ^{22}C_{22-19} = ^{22}C_3$.
$^{22}C_3 = \frac{22 \times 21 \times 20}{3 \times 2 \times 1} = 22 \times 7 \times 10 = 1540$.
Thus,there are $1540$ ways to arrange the books.

(C) Total number of points = $5 + 3 = 8$.
To form a triangle,we need to select $3$ points out of $8$,which can be done in $^8C_3$ ways.
However,if the $3$ selected points are collinear,they do not form a triangle.
There are $5$ points on one line,so the number of ways to select $3$ collinear points from these is $^5C_3$.
There are $3$ points on the other parallel line,so the number of ways to select $3$ collinear points from these is $^3C_3$.
Thus,the number of triangles = $^8C_3 - ^5C_3 - ^3C_3$.

(B) An octagon has $n = 8$ vertices.
To form a diagonal,we need to choose $2$ vertices out of $8$,which is given by $^8C_2$.
$^8C_2 = \frac{8 \times 7}{2 \times 1} = 28$.
These $28$ combinations include the $8$ sides of the octagon.
Therefore,the number of diagonals = $^8C_2 - 8 = 28 - 8 = 20$.

(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 $n$ must be a positive integer,$n = 11$.
Therefore,the number of sides is $11$.

(A) To form a triangle,we need to select $3$ non-collinear points.
Since all points lie on a circle,no three points are collinear.
Therefore,the number of triangles that can be formed by joining $4$ points is given by the combination formula $^n{C_r}$,where $n = 4$ and $r = 3$.
Number of triangles = $^4{C_3} = \frac{4!}{3!(4-3)!} = \frac{4 \times 3!}{3! \times 1!} = 4$.

(A) To form a triangle,we need to select $3$ points out of the given $9$ non-collinear points.
Since the points are non-collinear,any selection of $3$ points will form a unique triangle.
The number of ways to select $3$ points from $9$ is given by the combination formula $^nC_r = \frac{n!}{r!(n-r)!}$.
Substituting $n = 9$ and $r = 3$:
$^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$.
Therefore,$84$ triangles can be drawn.

(C) polygon with $m$ sides has $m$ vertices.
To form a line segment (side or diagonal),we choose $2$ vertices out of $m$,which is given by $^mC_2$.
Total number of line segments = $^mC_2 = \frac{m(m - 1)}{2}$.
These line segments include $m$ sides of the polygon.
Therefore,the number of diagonals = (Total line segments) - (Number of sides).
Number of diagonals = $\frac{m(m - 1)}{2} - m = \frac{m^2 - m - 2m}{2} = \frac{m(m - 3)}{2}$.

(D) To form a straight line,we need to select any $2$ points out of the given $8$ points.
Since the points are on a circle,no $3$ points are collinear.
Therefore,the number of straight lines is given by the combination formula $^nC_r = \frac{n!}{r!(n-r)!}$.
Here,$n = 8$ and $r = 2$.
Number of lines = $^8C_2 = \frac{8 \times 7}{2 \times 1} = 28$.

(A) To form a triangle,we need to select $3$ non-collinear points from the given set of points.
Total number of ways to select $3$ points from $12$ points is given by $^{12}C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
Since $7$ points are collinear,selecting any $3$ points from these $7$ points will not form a triangle.
The number of ways to select $3$ points from these $7$ collinear points is $^{7}C_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.
Therefore,the number of triangles that can be formed is the total number of selections minus the number of collinear selections:
Number of triangles $= 220 - 35 = 185$.

(B) To form a triangle,we need $3$ non-collinear points.
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 $4$ points are collinear,selecting these $4$ points will not form a triangle. The number of ways to select $3$ points from these $4$ collinear points is $^4C_3$.
$^4C_3 = 4$.
Therefore,the number of triangles that can be formed is $120 - 4 = 116$.

(A) The total number of ways to select $2$ points out of $16$ is given by $^{16}C_2 = \frac{16 \times 15}{2} = 120$.
Since $6$ points are collinear,the number of lines formed by these $6$ points is only $1$,instead of $^{6}C_2 = \frac{6 \times 5}{2} = 15$.
Therefore,the total number of lines that can be drawn is $^{16}C_2 - ^{6}C_2 + 1 = 120 - 15 + 1 = 106$.

(B) The total number of points is $m + n + k$. The number of ways to select $3$ points from these is $^{m+n+k}C_3$.
However,selecting $3$ points from the same line does not form a triangle. The number of such collinear sets is $^mC_3 + ^nC_3 + ^kC_3$.
Therefore,the required number of triangles is $^{m+n+k}C_3 - ^mC_3 - ^nC_3 - ^kC_3$.

(B) To form a parallelogram,we need to select $2$ lines from the first set of $4$ parallel lines and $2$ lines from the second set of $3$ parallel lines.
The number of ways to select $2$ lines from $4$ is given by $^{4}C_{2} = \frac{4 \times 3}{2 \times 1} = 6$.
The number of ways to select $2$ lines from $3$ is given by $^{3}C_{2} = \frac{3 \times 2}{2 \times 1} = 3$.
The total number of parallelograms is the product of these two combinations:
Total $= ^{4}C_{2} \times ^{3}C_{2} = 6 \times 3 = 18$.

(C) The number of lines formed by joining $6$ points is $^6C_2 = \frac{6 \times 5}{2} = 15$.
The total number of intersection points of these $15$ lines,assuming no three lines are concurrent,is $^{15}C_2 = \frac{15 \times 14}{2} = 105$.
However,at each of the original $6$ points,$5$ lines intersect. Since these $5$ lines are counted as intersecting at a single point,we must subtract the extra counts.
For each of the $6$ points,the number of pairs of lines intersecting at that point is $^5C_2 = 10$.
Since these $10$ pairs are counted as $10$ distinct intersection points in the initial $105$,but they all represent only $1$ point,we subtract $10 - 1 = 9$ for each of the $6$ points.
Total distinct points of intersection $= 105 - 6 \times (10 - 1) = 105 - 54 = 51$.

(A) Case $I$: When $A$ is excluded.
Number of triangles = (selection of $2$ points from $AB$ and $1$ point from $AC$) + (selection of $1$ point from $AB$ and $2$ points from $AC$).
Number of triangles = $^mC_2 \times ^nC_1 + ^mC_1 \times ^nC_2 = \frac{m(m-1)}{2} \times n + m \times \frac{n(n-1)}{2} = \frac{mn}{2}(m-1+n-1) = \frac{mn(m+n-2)}{2}$.
Case $II$: When $A$ is included.
Triangles can be formed by taking $A$ as one vertex and two other points from the lines $AB$ and $AC$.
If we take $A$ as a vertex,we need one point from $AB$ and one point from $AC$ to form a triangle.
Number of triangles = $^mC_1 \times ^nC_1 = mn$.
Total triangles when $A$ is included = (Triangles not involving $A$) + (Triangles involving $A$) = $\frac{mn(m+n-2)}{2} + mn = \frac{mn}{2}(m+n-2+2) = \frac{mn(m+n)}{2}$.
Required ratio = $\frac{\text{Case } I}{\text{Case } II} = \frac{\frac{mn(m+n-2)}{2}}{\frac{mn(m+n)}{2}} = \frac{m+n-2}{m+n}$.

(C) Since no two lines are parallel and no three are concurrent,the $n$ straight lines intersect at $N = ^nC_2$ points.
Two points are required to determine a straight line,so the total number of lines obtained by joining these $N$ points is $^NC_2$.
However,each original line contains $(n - 1)$ intersection points. The number of lines formed by these $(n - 1)$ points on a single original line is $^{n-1}C_2$. Since there are $n$ such original lines,we must subtract these lines from the total.
Thus,the number of fresh lines is $^NC_2 - n \times ^{n-1}C_2$.
Substituting $N = \frac{n(n-1)}{2}$,we get:
$\frac{N(N-1)}{2} - \frac{n(n-1)(n-2)}{2} = \frac{\frac{n(n-1)}{2} (\frac{n(n-1)}{2} - 1)}{2} - \frac{n(n-1)(n-2)}{2}$
$= \frac{n(n-1)(n^2-n-2)}{8} - \frac{4n(n-1)(n-2)}{8}$
$= \frac{n(n-1)(n-2)(n+1) - 4n(n-1)(n-2)}{8}$
$= \frac{n(n-1)(n-2)(n-3)}{8}$.

(C) parallelogram is defined by two pairs of parallel lines.
Initially,the parallelogram has $2$ sides in each direction.
When $m$ lines are added parallel to each set of sides,each set now contains $m + 2$ parallel lines.
$A$ parallelogram is formed by selecting $2$ lines from the first set of $m + 2$ lines and $2$ lines from the second set of $m + 2$ lines.
The number of ways to choose $2$ lines from $m + 2$ lines is given by $^{m+2}C_2$.
Since we need to choose from both sets independently,the total number of parallelograms is $^{m+2}C_2 \times {^{m+2}C_2} = ({^{m+2}C_2})^2$.

(A) The total number of intersection points for $37$ lines,if no three were concurrent and no two were parallel,would be $^{37}C_2 = \frac{37 \times 36}{2} = 666$.
However,$13$ lines pass through point $A$. These lines should have produced $^{13}C_2 = \frac{13 \times 12}{2} = 78$ intersection points,but they only produce $1$ point. Thus,we must subtract $78$ and add $1$.
Similarly,$11$ lines pass through point $B$. These lines should have produced $^{11}C_2 = \frac{11 \times 10}{2} = 55$ intersection points,but they only produce $1$ point. Thus,we must subtract $55$ and add $1$.
The total number of intersection points is $666 - 78 + 1 - 55 + 1 = 535$.

(D) The maximum number of intersection points is calculated as follows:
$1$. Intersection between $8$ straight lines: $^8C_2 = \frac{8 \times 7}{2} = 28$.
$2$. Intersection between $4$ circles: $^4C_2 \times 2 = 6 \times 2 = 12$ (since two circles intersect at $2$ points).
$3$. Intersection between $8$ straight lines and $4$ circles: $^8C_1 \times ^4C_1 \times 2 = 8 \times 4 \times 2 = 64$ (since a line and a circle intersect at $2$ points).
Total points $= 28 + 12 + 64 = 104$.

(C) Given $18$ points in a plane,where $5$ points are collinear.
$(i)$ Number of straight lines:
Total lines possible from $18$ points is $^{18}C_2$.
Since $5$ points are collinear,they form only $1$ line instead of $^5C_2$ lines.
Number of lines $= ^{18}C_2 - ^5C_2 + 1 = \frac{18 \times 17}{2} - \frac{5 \times 4}{2} + 1 = 153 - 10 + 1 = 144$.
$(ii)$ Number of triangles:
Total triangles possible from $18$ points is $^{18}C_3$.
Since $5$ points are collinear,they do not form any triangle.
Number of triangles $= ^{18}C_3 - ^5C_3 = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} - \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 816 - 10 = 806$.

(A) To form a triangle,we need to select $3$ non-collinear points from the given $16$ points.
Total ways to select $3$ points from $16$ is given by $^{16}C_3 = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560$.
Since $8$ points are collinear,any selection of $3$ points from these $8$ points will not form a triangle.
The number of ways to select $3$ points from these $8$ collinear points is $^{8}C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
Therefore,the number of triangles that can be formed is $560 - 56 = 504$.

(B) The number of triangles formed by $n$ vertices is given by $T_n = ^nC_3$.
Given the equation $T_{n+1} - T_n = 21$,we substitute the formula:
$^{n+1}C_3 - ^nC_3 = 21$.
Using the Pascal identity property,$^nC_r + ^nC_{r-1} = ^{n+1}C_r$,we can write $^{n+1}C_3 = ^nC_3 + ^nC_2$.
Substituting this into the equation:
$(^nC_3 + ^nC_2) - ^nC_3 = 21$
$^nC_2 = 21$.
Expanding the combination formula:
$\frac{n(n-1)}{2} = 21$
$n(n-1) = 42$
$n(n-1) = 7 \times 6$.
Thus,$n = 7$.

(A) To form a triangle,we need to select $3$ non-collinear points from the given set of points.
Total number of ways to select $3$ points from $10$ points is $^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
Since $6$ points are collinear,any selection of $3$ points from these $6$ points will not form a triangle.
The number of ways to select $3$ points from these $6$ collinear points is $^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
Therefore,the number of triangles formed = $^{10}C_3 - ^6C_3 = 120 - 20 = 100$.

(C) Total number of points is $n$. The number of ways to select $2$ points from $n$ points is $^{n}C_{2}$.
Since $p$ points are collinear,they all lie on a single line. The number of ways to select $2$ points from these $p$ points is $^{p}C_{2}$.
These $^{p}C_{2}$ selections would normally represent distinct lines,but since the points are collinear,they all form only $1$ single line.
Therefore,the total number of lines is given by the formula: $^{n}C_{2} - ^{p}C_{2} + 1$.

(A) The total number of ways to select $3$ line segments out of $6$ is given by $^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
$A$ triangle can be formed if the sum of the lengths of any two sides is greater than the third side.
Let the set of lengths be $S = \{2, 3, 4, 5, 6, 7\}$.
The combinations that do $NOT$ form a triangle are those where the sum of the two smaller sides is less than or equal to the largest side:
$1$. $(2, 3, 5)$ since $2+3=5$
$2$. $(2, 3, 6)$ since $2+3 < 6$
$3$. $(2, 3, 7)$ since $2+3 < 7$
$4$. $(2, 4, 6)$ since $2+4=6$
$5$. $(2, 4, 7)$ since $2+4 < 7$
$6$. $(2, 5, 7)$ since $2+5=7$
$7$. $(3, 4, 7)$ since $3+4=7$
There are $7$ such combinations.
Therefore,the number of triangles that can be formed is $20 - 7 = 13$.
Looking at the options,the expression $^6C_3 - 7$ equals $20 - 7 = 13$.

(C) 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 $35$,we have:
$\frac{n(n-3)}{2} = 35$
$n(n-3) = 70$
$n^2 - 3n - 70 = 0$
Factoring the quadratic equation:
$(n - 10)(n + 7) = 0$
Since $n$ must be a positive integer,we have $n = 10$.
Therefore,the number of sides is $10$.

(D) The total number of lines formed by $n$ points is given by $^{n}C_{2}$.
Given $n = 20$ points,the total number of lines without any collinearity constraint is $^{20}C_{2} = \frac{20 \times 19}{2} = 190$.
Since $4$ points are collinear,they form only $1$ line instead of $^{4}C_{2} = \frac{4 \times 3}{2} = 6$ lines.
Therefore,the required number of lines is $190 - 6 + 1 = 185$.

(D) rectangle is formed by choosing two horizontal lines and two vertical lines.
Let the horizontal lines be $y_0, y_1, \dots, y_{2n-1}$ and vertical lines be $x_0, x_1, \dots, x_{2m-1}$.
$A$ rectangle has odd side lengths if the difference between the indices of the chosen lines is odd.
For the horizontal side,we need to choose two lines $y_i, y_j$ such that $|i - j|$ is odd. This means one index must be even and the other must be odd.
In the set ${0, 1, \dots, 2n-1}$,there are $n$ even numbers ${0, 2, \dots, 2n-2}$ and $n$ odd numbers ${1, 3, \dots, 2n-1}$.
The number of ways to choose one even and one odd index is $n \times n = n^2$.
Similarly,for the vertical side,the number of ways to choose two lines such that the difference is odd is $m \times m = m^2$.
Therefore,the total number of rectangles with odd side lengths is $m^2 \times n^2 = m^2n^2$.

(A) First,arrange $m$ men in a row in $m!$ ways.
Since $n < m$ and no two women can sit together,we consider the gaps between the men.
In any one of the $m!$ arrangements,there are $(m + 1)$ available spaces (including the ends) where $n$ women can be placed.
The number of ways to arrange $n$ women in these $(m + 1)$ spaces is given by $^{m+1}P_n$.
Therefore,by the fundamental principle of counting,the total number of arrangements is $m! \times {}^{m+1}P_n$.
$= m! \times \frac{(m + 1)!}{(m + 1 - n)!} = \frac{m! (m + 1)!}{(m - n + 1)!}$.

(C) Let the number of men be $n$. The total number of participants is $n + 2$.
Each participant plays $2$ games with every other participant.
The number of games played between men is $2 \times {^nC_2} = 2 \times \frac{n(n-1)}{2} = n(n-1) = n^2 - n$.
The number of games played between men and women is $2 \times (n \times 2) = 4n$.
According to the problem,the number of games between men exceeds the number of games between men and women by $66$:
$n^2 - n - 4n = 66$
$n^2 - 5n - 66 = 0$
$(n - 11)(n + 6) = 0$
Since $n$ must be positive,$n = 11$.
The total number of participants is $n + 2 = 11 + 2 = 13$.

(B) There are $10$ students in total,including $3$ girls and $7$ boys.
To ensure no two girls are seated together,we first arrange the $7$ boys in a row,which can be done in $7!$ ways.
This creates $8$ possible gaps (including the ends) where the $3$ girls can be seated: $\_ B_1 \_ B_2 \_ B_3 \_ B_4 \_ B_5 \_ B_6 \_ B_7 \_$.
The number of ways to choose $3$ gaps out of $8$ and arrange the $3$ girls in them is given by $^8P_3$.
Therefore,the total number of arrangements is $7! \times ^8P_3$.

(C) The total number of ways to choose $3$ vertices out of $6$ is given by the combination formula ${}^6C_3 = \frac{6 \times 5 \times 4}{1 \times 2 \times 3} = 20$.
In a regular hexagon,there are exactly $2$ equilateral triangles that can be formed by connecting the vertices. These are formed by taking alternating vertices (e.g.,vertices $1, 3, 5$ and $2, 4, 6$).
Therefore,the required probability is $\frac{\text{Number of equilateral triangles}}{\text{Total number of triangles}} = \frac{2}{20} = \frac{1}{10}$.

(B) Total number of ways to arrange $6$ boys and $6$ girls in a row is $12!$.
To find the number of arrangements where all $6$ girls sit together,we treat the $6$ girls as a single unit. This unit along with the $6$ boys gives us $7$ entities,which can be arranged in $7!$ ways.
The $6$ girls within their unit can be arranged among themselves in $6!$ ways.
So,the number of favorable arrangements is $7! \times 6!$.
The required probability is $\frac{7! \times 6!}{12!} = \frac{7! \times 720}{12 \times 11 \times 10 \times 9 \times 8 \times 7!} = \frac{720}{95040} = \frac{1}{132}$.

(B) The total number of ways to arrange $7$ white balls and $3$ black balls in a row is given by the combination formula $\binom{10}{3} = \frac{10!}{7!3!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
To ensure no two black balls are adjacent,we first arrange the $7$ white balls in a row. This creates $8$ possible spaces (gaps) where the $3$ black balls can be placed (one before the first ball,one after the last ball,and $6$ between the balls).
The number of ways to choose $3$ spaces out of $8$ is $\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
Therefore,the required probability is $\frac{56}{120} = \frac{7}{15}$.

(B) The total number of points is $m + n + k$.
Total ways to select $3$ points from these is $^{m+n+k}C_3$.
However,points lying on the same line cannot form a triangle.
Number of triangles formed by $3$ points on line $l_1$ is $^mC_3$.
Number of triangles formed by $3$ points on line $l_2$ is $^nC_3$.
Number of triangles formed by $3$ points on line $l_3$ is $^kC_3$.
Therefore,the number of triangles that can be formed is $^{m+n+k}C_3 - ^mC_3 - ^nC_3 - ^kC_3$.

(B) To form a parallelogram,we need to select $2$ lines from the set of $5$ parallel lines and $2$ lines from the set of $4$ parallel lines.
The number of ways to select $2$ lines from $5$ is given by $^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$.
The number of ways to select $2$ lines from $4$ is given by $^4C_2 = \frac{4 \times 3}{2 \times 1} = 6$.
Therefore,the total number of parallelograms formed is $^5C_2 \times ^4C_2 = 10 \times 6 = 60$.

(C) The number of ways to choose $2$ vertices from $n$ vertices is given by $\binom{n}{2}$.
This includes the $n$ sides of the polygon.
To find the number of diagonals,we subtract the number of sides from the total number of lines formed by connecting any two vertices:
$\text{Number of diagonals} = \binom{n}{2} - n$
$= \frac{n(n - 1)}{2} - n$
$= \frac{n^2 - n - 2n}{2}$
$= \frac{n^2 - 3n}{2}$
$= \frac{1}{2}n(n - 3)$

(B) straight line is formed by joining any two points. Therefore,the total number of lines that can be formed by joining $10$ points is $^{10}C_2$.
However,$4$ of these points are collinear. Joining any two of these $4$ points results in the same line.
The number of lines formed by $4$ collinear points is $^{4}C_2$,which counts as only $1$ line instead of $6$.
Therefore,the total number of distinct lines = $^{10}C_2 - ^{4}C_2 + 1$.
Calculation: $\frac{10 \times 9}{2} - \frac{4 \times 3}{2} + 1 = 45 - 6 + 1 = 40$.

(B) Given that $T_n = \binom{n}{3}$.
We are given $T_{n+1} - T_n = 21$.
Using the property $\binom{n+1}{r} = \binom{n}{r} + \binom{n}{r-1}$,we have:
$T_{n+1} = \binom{n+1}{3} = \binom{n}{3} + \binom{n}{2}$.
Substituting this into the equation:
$(\binom{n}{3} + \binom{n}{2}) - \binom{n}{3} = 21$.
$\binom{n}{2} = 21$.
$\frac{n(n-1)}{2} = 21$.
$n(n-1) = 42$.
$n(n-1) = 7 \times 6$.
Thus,$n = 7$.

(B) Step $1$: Select $4$ glass beads from $6$ available: $^6C_4 = 15$ ways.
Step $2$: Select $4$ metal beads from $5$ available: $^5C_4 = 5$ ways.
Step $3$: Total beads selected = $4 + 4 = 8$ beads.
Step $4$: The number of ways to arrange $n$ distinct beads in a necklace is $\frac{(n-1)!}{2}$.
Step $5$: For $8$ beads,the number of arrangements is $\frac{(8-1)!}{2} = \frac{7!}{2!}$.
Step $6$: Total number of necklaces = $^6C_4 \times ^5C_4 \times \frac{7!}{2!}$.

(B) An octagon has $8$ vertices.
The total number of line segments formed by joining any two vertices is given by $^nC_2$,where $n = 8$.
Total line segments $= ^8C_2 = \frac{8 \times 7}{2} = 28$.
Out of these $28$ line segments,$8$ are the sides of the octagon.
The number of diagonals is the total number of line segments minus the number of sides.
Number of diagonals $= 28 - 8 = 20$.

(B) The number of triangles that can be formed by joining the vertices of a regular polygon with $n$ sides is given by the combination formula $T_n = \binom{n}{3} = \frac{n(n-1)(n-2)}{6}$.
Given the condition $T_{(n+1)} - T_n = 10$,we substitute the formula:
$\binom{n+1}{3} - \binom{n}{3} = 10$.
Using the identity $\binom{n+1}{k} - \binom{n}{k} = \binom{n}{k-1}$,we get:
$\binom{n}{2} = 10$.
$\frac{n(n-1)}{2} = 10$.
$n(n-1) = 20$.
$n^2 - n - 20 = 0$.
$(n-5)(n+4) = 0$.
Since $n$ must be a positive integer,$n = 5$.

(A) For a triangle to be formed,the sum of the lengths of any two sides must be strictly greater than the length of the third side.
The total number of ways to select $3$ segments out of $6$ is $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
We must subtract the combinations that do not satisfy the triangle inequality $(a + b \leq c)$:
$(2, 3, 5), (2, 3, 6), (2, 3, 7), (2, 4, 6), (2, 4, 7), (2, 5, 7), (3, 4, 7)$.
There are $7$ such combinations that cannot form a triangle.
Therefore,the number of triangles that can be formed is $\binom{6}{3} - 7$.

(D) To form a line,we need $2$ points. The total number of ways to select $2$ points from $20$ is given by $\binom{20}{2}$.
Since $4$ points are collinear,the number of lines formed by these $4$ points is only $1$ instead of $\binom{4}{2}$.
The formula for the number of lines is $\binom{n}{2} - \binom{m}{2} + 1$,where $n$ is the total number of points and $m$ is the number of collinear points.
Here,$n = 20$ and $m = 4$.
Number of lines $= \binom{20}{2} - \binom{4}{2} + 1$
$= \frac{20 \times 19}{2} - \frac{4 \times 3}{2} + 1$
$= 190 - 6 + 1 = 185$.

(D) triangle is formed by selecting $3$ non-collinear points.
Total points on $l_1$ (excluding $P$) $= 3$.
Total points on $l_2$ (excluding $P$) $= 5$.
Total points $= 3 + 5 = 8$.
To form a triangle,we can select:
$1$. $2$ points from $l_1$ and $1$ point from $l_2$: $\binom{3}{2} \times \binom{5}{1} = 3 \times 5 = 15$.
$2$. $1$ point from $l_1$ and $2$ points from $l_2$: $\binom{3}{1} \times \binom{5}{2} = 3 \times 10 = 30$.
Total triangles $= 15 + 30 = 45$.