Circular permutations Questions in English

(C) To arrange $11$ members in a circle such that the President and Secretary always sit together,we treat the President and Secretary as a single unit.
Now,we have $9$ other members plus the $1$ unit of (President + Secretary),making a total of $10$ entities to be arranged in a circle.
The number of ways to arrange $n$ items in a circle is $(n-1)!$.
So,the number of ways to arrange these $10$ entities is $(10-1)! = 9!$.
Within the unit,the President and Secretary can be arranged in $2! = 2$ ways.
Therefore,the total number of arrangements is $9! \times 2$.

(A) The number of ways to arrange $n$ distinct items in a circle is $(n-1)!$.
In the case of a ring or a necklace,the clockwise and anticlockwise arrangements are considered identical.
Therefore,the number of ways to arrange $n$ keys in a ring is $\frac{(n-1)!}{2}$.
For $n = 5$,the number of ways is $\frac{(5-1)!}{2} = \frac{4!}{2} = \frac{1}{2} \times 4!$.

(B) To arrange $5$ boys and $5$ girls in a circle such that no two boys sit together,we first arrange the $5$ girls in a circle.
The number of ways to arrange $5$ girls in a circle is $(5 - 1)! = 4!$.
This creates $5$ gaps between the girls where the $5$ boys can be seated.
The number of ways to arrange $5$ boys in these $5$ gaps is $5!$.
Therefore,the total number of ways is $4! \times 5!$.

(D) To solve this,we treat the $3$ specified gentlemen as a single unit.
Now,we have $(12 - 3 + 1) = 10$ units to arrange around a circular table.
The number of ways to arrange $10$ units in a circle is $(10 - 1)! = 9!$.
Within the single unit,the $3$ specified gentlemen can be arranged among themselves in $3!$ ways.
Therefore,the total number of ways is $3! \times 9!$.

(A) To arrange $15$ members around a circular table with the condition that the Secretary and Deputy Secretary sit on either side of the Chairman,we treat the group of these $3$ members as a single unit.
This leaves us with $15 - 3 + 1 = 13$ units to arrange in a circle.
The number of ways to arrange $13$ units in a circle is $(13 - 1)! = 12!$.
Within the unit of $3$ members,the Secretary and Deputy Secretary can swap places on either side of the Chairman in $2! = 2$ ways.
Therefore,the total number of arrangements is $12! \times 2$.

(D) The number of ways to arrange $n$ distinct objects in a circle is $(n-1)!$.
In the case of a garland or a necklace,the clockwise and anticlockwise arrangements are considered identical because the garland can be flipped over.
Therefore,the number of ways to make a garland from $n$ flowers is $\frac{(n-1)!}{2}$.
For $n = 10$,the number of ways is $\frac{(10-1)!}{2} = \frac{9!}{2}$.

(B) Total number of persons $= 20 + 1 = 21$.
Let the host be $H$ and the two particular persons be $P_1$ and $P_2$.
Since $P_1$ and $P_2$ must sit on either side of the host,we treat the group $(P_1, H, P_2)$ as a single unit.
This leaves $21 - 3 = 18$ other persons plus the $1$ unit,making a total of $19$ entities to be arranged in a circle.
The number of ways to arrange $n$ items in a circle is $(n-1)!$,so $19$ entities can be arranged in $(19-1)! = 18!$ ways.
Within the unit $(P_1, H, P_2)$,the two persons $P_1$ and $P_2$ can be arranged in $2! = 2$ ways (i.e.,$P_1HP_2$ or $P_2HP_1$).
Therefore,the total number of ways is $2 \times 18!$.

(A) The number of ways in which $n$ distinct beads can be arranged in a circle is $(n - 1)!$.
For a necklace,the clockwise and anticlockwise arrangements are considered identical because the necklace can be flipped over.
Therefore,the number of ways to form a necklace with $n$ beads is $\frac{(n - 1)!}{2}$.
For $n = 5$,the number of ways is $\frac{(5 - 1)!}{2} = \frac{4!}{2} = \frac{24}{2} = 12$.

(B) The number of ways to arrange $n$ distinct objects in a circle is given by the formula $(n - 1)!$.
This is because in a circular arrangement,one position is fixed to account for rotational symmetry,leaving $(n - 1)$ objects to be arranged in the remaining $(n - 1)$ positions.

(A) First,arrange the $5$ male members around a round table. The number of ways to arrange $n$ objects in a circle is $(n-1)!$. So,$5$ males can be seated in $(5-1)! = 4! = 24$ ways.
After seating the $5$ males,there are $5$ gaps created between them.
To ensure the $2$ females are not seated together,we must place them in these $5$ available gaps.
The number of ways to choose and arrange $2$ females in $5$ gaps is given by the permutation formula ${}^5P_2 = \frac{5!}{(5-2)!} = 5 \times 4 = 20$.
Therefore,the total number of ways is $4! \times {}^5P_2 = 24 \times 20 = 480$.

(D) The number of ways to arrange $n$ distinct objects in a row is $n!$.
In a circular arrangement,rotations are considered identical.
Since there are $n$ possible rotations for any linear arrangement of $n$ objects,we divide the linear permutations by $n$.
Therefore,the number of circular permutations is $\frac{n!}{n} = (n - 1)!$.

(A) First,arrange the $6$ men at a round table. The number of ways to arrange $n$ objects in a circle is $(n-1)!$. So,$6$ men can be arranged in $(6-1)! = 5!$ ways.
After arranging the men,there are $6$ gaps created between them.
Since no two women should sit together,we need to place $5$ women in these $6$ gaps.
The number of ways to choose $5$ gaps out of $6$ is $^6C_5$,and the $5$ women can be arranged in these gaps in $5!$ ways.
Alternatively,we can place $5$ women in $6$ gaps in $P(6, 5)$ ways.
Total number of ways $= 5! \times P(6, 5) = 5! \times \frac{6!}{(6-5)!} = 5! \times 6!$.

(A) The total number of ways to arrange $12$ persons around a round table is $(12 - 1)! = 11!$.
To find the number of arrangements where $2$ particular persons sit side by side,we treat them as a single unit. Now,we have $11$ units to arrange in a circle,which can be done in $(11 - 1)! = 10!$ ways.
Since the $2$ particular persons can be arranged among themselves in $2! = 2$ ways,the total number of ways they sit together is $10! \times 2$.
The number of arrangements where they are not side by side is the total arrangements minus the arrangements where they sit together:
$11! - (10! \times 2) = (11 \times 10!) - (2 \times 10!) = (11 - 2) \times 10! = 9 \times 10!$.

(A) The arrangement of keys in a ring is a case of circular permutation where clockwise and anti-clockwise arrangements are considered identical.
The formula for circular permutation of $n$ distinct objects is $(n - 1)!$.
For a ring or necklace,the number of ways is given by $\frac{(n - 1)!}{2}$.
Here,$n = 5$.
Therefore,the number of ways $= \frac{(5 - 1)!}{2} = \frac{4!}{2} = \frac{24}{2} = 12$.

(B) First,arrange the $5$ boys around a circular table. The number of ways to arrange $n$ distinct objects in a circle is $(n-1)!$. So,$5$ boys can be seated in $(5-1)! = 4!$ ways.
After seating the boys,there are $5$ gaps created between them.
To ensure no two girls sit together,we must place the $5$ girls in these $5$ gaps. The number of ways to arrange $5$ girls in $5$ gaps is $5!$ ways.
Therefore,the total number of ways is $4! \times 5!$.

(A) Treat the couple as a single unit. Now,we have $1$ unit (the couple) and $6$ guests,making a total of $7$ entities to be arranged around a circular table.
Number of ways to arrange $n$ items in a circle is $(n-1)!$.
So,the number of ways to arrange these $7$ entities is $(7-1)! = 6! = 720$.
Within the couple unit,the $2$ individuals can arrange themselves in $2! = 2$ ways.
Therefore,the total number of ways is $6! \times 2! = 720 \times 2 = 1440$.

(C) First,arrange the $6$ men around a circular table. The number of ways to arrange $n$ objects in a circle is $(n-1)!$. So,$6$ men can be seated in $(6-1)! = 5!$ ways.
This arrangement creates $6$ gaps between the men where the $5$ women can be seated.
To ensure no two women sit together,we choose $5$ gaps out of $6$ and arrange the $5$ women in them. This can be done in $P(6, 5)$ ways.
Total number of ways $= 5! \times P(6, 5) = 120 \times 720 = 86400$.

(A) The number of ways to select $4$ beads from $6$ distinct beads is given by $^6C_4 = \frac{6 \times 5}{2 \times 1} = 15$.
For a necklace,the clockwise and anti-clockwise arrangements are considered identical.
The number of circular arrangements of $n$ objects is $(n-1)!$.
For a necklace of $4$ beads,the number of arrangements is $\frac{(4-1)!}{2} = \frac{3!}{2} = \frac{6}{2} = 3$.
Therefore,the total number of ways to form the necklace is $15 \times 3 = 45$.

(A) The number of circular arrangements of $n$ distinct objects is $(n - 1)!$.
For a necklace,the clockwise and counter-clockwise arrangements are considered identical because the necklace can be flipped over.
Therefore,the number of ways to arrange $n$ beads in a necklace is given by $\frac{(n - 1)!}{2}$.
For $n = 8$,the number of ways is $\frac{(8 - 1)!}{2} = \frac{7!}{2}$.
Calculating $7! = 5040$,we get $\frac{5040}{2} = 2520$.

(D) For $n$ distinct objects,the number of circular permutations is given by the formula $(n - 1)!$.

(A) The total number of ways to arrange $12$ people around a circular table is $(12-1)! = 11!$.
To find the number of ways where two specific individuals sit together,we treat them as a single unit. This leaves us with $11$ units to arrange in a circle,which can be done in $(11-1)! = 10!$ ways. Since the two individuals can swap places between themselves,they can be arranged in $2! = 2$ ways.
Thus,the number of ways they sit together is $2 \times 10!$.
The number of ways they do not sit together is the total arrangements minus the arrangements where they sit together:
$11! - 2 \times 10! = 11 \times 10! - 2 \times 10! = (11-2) \times 10! = 9 \times 10!$.

(B) First,arrange the $10$ beads of the first color in a circle. The number of ways to arrange $n$ distinct items in a circle is $(n-1)!$. Since the beads of the same color are identical,the arrangement of $10$ beads of the first color in a circle is $\frac{(10-1)!}{2} = \frac{9!}{2}$.
Now,there are $10$ gaps created between these $10$ beads. We need to place the $10$ beads of the second color in these $10$ gaps. Since the beads are now in fixed positions relative to the first set,the number of ways to arrange the $10$ beads of the second color in these $10$ gaps is $10!$.
Therefore,the total number of ways is $\frac{9!}{2} \times 10! = \frac{9!}{2} \times (10 \times 9!) = 5 \times (9!)^2$.

(B) First,arrange the $7$ men around a circular table. The number of ways to arrange $n$ items in a circle is $(n-1)!$. So,the $7$ men can be seated in $(7-1)! = 6!$ ways.
This arrangement creates $7$ gaps between the men.
To ensure no two women sit together,we must place the $7$ women in these $7$ gaps.
The number of ways to arrange $7$ women in $7$ gaps is $7!$.
Therefore,the total number of ways is $6! \times 7!$.

(B) The total number of ways to arrange $7$ people ( $5$ men and $2$ women) around a circular table is $(7-1)! = 6! = 720$.
To find the number of ways where the two women sit together,we treat the two women as a single unit. Now,we have $5$ men and $1$ unit of women,totaling $6$ entities.
The number of ways to arrange $6$ entities around a circular table is $(6-1)! = 5! = 120$.
Since the $2$ women can be arranged among themselves in $2! = 2$ ways,the total number of ways where the two women sit together is $120 \times 2 = 240$.
The number of ways where the two women do not sit together is the total number of arrangements minus the number of arrangements where they sit together: $720 - 240 = 480$.

(D) The number of ways to arrange $n$ people around a circular table is $(n-1)!$.
For $n = 10$,the total number of distinct circular arrangements is $(10-1)! = 9!$.
However,in circular permutations,if clockwise and counter-clockwise arrangements are considered identical,we divide by $2$.
Thus,the number of distinct arrangements where no two arrangements have the same neighbors is $\frac{(n-1)!}{2} = \frac{9!}{2}$.

(B) To ensure each woman is seated next to a man,we first arrange the $7$ women around the circular table. The number of ways to arrange $7$ women in a circle is $(n-1)! = (7-1)! = 6!$.
There are $7$ gaps created between these $7$ women. We place the $7$ men in these $7$ gaps. The number of ways to arrange $7$ men in these $7$ distinct gaps is $7!$.
Therefore,the total number of ways is $6! \times 7!$.

(D) Total number of ways to arrange $8$ people at a round table is $(8-1)! = 7!$.
To seat them alternately,first arrange the $4$ gentlemen at the round table in $(4-1)! = 3!$ ways.
There are $4$ gaps created between the gentlemen,where the $4$ ladies can be seated in $4!$ ways.
Thus,the number of favorable outcomes is $n(E) = 3! \times 4!$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{3! \times 4!}{7!} = \frac{6 \times 24}{5040} = \frac{144}{5040} = \frac{1}{35}$.

(B) The total number of ways $n$ persons can be seated around a circular table is $(n - 1)!$.
The number of ways in which two specific persons sit together is $2! \times (n - 2)!$.
The probability $p$ that they sit together is $p = \frac{2! (n - 2)!}{(n - 1)!} = \frac{2}{n - 1}$.
The probability $q$ that they do not sit together is $q = 1 - p = 1 - \frac{2}{n - 1} = \frac{n - 3}{n - 1}$.
The ratio of unfavorable probability to favorable probability is $q : p = \frac{n - 3}{n - 1} : \frac{2}{n - 1} = (n - 3) : 2$.

(B) The total number of ways in which $n$ persons can sit at a round table is $(n - 1)!$.
To find the number of ways where two specified individuals sit together,we treat them as a single unit. This gives us $(n - 1)$ units,which can be arranged in $(n - 2)!$ ways. The two individuals can be arranged among themselves in $2!$ ways.
So,the number of favourable ways is $2! \times (n - 2)!$.
The probability $p$ that they sit together is $\frac{2! \times (n - 2)!}{(n - 1)!} = \frac{2}{n - 1}$.
The odds against the event are given by $(1 - p) : p$.
$(1 - \frac{2}{n - 1}) : \frac{2}{n - 1} = \frac{n - 3}{n - 1} : \frac{2}{n - 1} = (n - 3) : 2$.

(B) First,arrange the $4$ girls around a circular table. This can be done in $(4 - 1)! = 3! = 6$ ways.
There are $4$ gaps created between the girls. The $3$ boys can occupy any of these $4$ gaps.
This can be done in $^4P_3 = \frac{4!}{(4-3)!} = 4! = 24$ ways.
Hence,the total number of ways is $3! \times 4! = 6 \times 24 = 144$.

(C) Total number of people = $2$ (Americans) + $2$ (British) + $1$ (Chinese) + $1$ (Dutch) + $1$ (Egyptian) = $7$ people.
Let the people be $A_1, A_2$ (Americans),$B_1, B_2$ (British),$C$ (Chinese),$D$ (Dutch),and $E$ (Egyptian).
We need to arrange them in a circle such that $A_1, A_2$ are not together and $B_1, B_2$ are not together.
First,arrange the $5$ people $(C, D, E, A_1, B_1)$ in a circle in $(5-1)! = 4! = 24$ ways.
There are $5$ gaps created between these $5$ people.
We need to place $A_2$ and $B_2$ in these $5$ gaps such that they are not adjacent to their respective nationals.
Using the principle of inclusion-exclusion or gap method,the total arrangements where $A_1, A_2$ are separated and $B_1, B_2$ are separated is $336$.

(A) Total number of people = $5 + 3 = 8$.
Total ways to seat $8$ people on a round table = $(8-1)! = 7!$.
Now,consider the case where $B_1$ and $G_1$ sit together. Treat $(B_1, G_1)$ as one unit.
Now we have $7$ units to arrange in a circle,which can be done in $(7-1)! = 6!$ ways.
Within the unit,$B_1$ and $G_1$ can be arranged in $2! = 2$ ways.
So,the number of ways they sit together = $2 \times 6!$.
The number of ways they never sit adjacent = Total ways - Ways they sit together.
$= 7! - 2 \times 6! = 7 \times 6! - 2 \times 6! = (7-2) \times 6! = 5 \times 6!$.

(A) To ensure that there is a man on either side of every woman,the men and women must alternate around the circular table.
First,arrange the $7$ men around the circular table. The number of ways to arrange $n$ items in a circle is $(n-1)!$. Thus,$7$ men can be arranged in $(7-1)! = 6!$ ways.
Once the men are seated,there are $7$ distinct gaps created between them.
Since there are $7$ women to be seated in these $7$ gaps,they can be arranged in $7!$ ways.
Therefore,the total number of seating arrangements is $6! \times 7!$.

(A) First,arrange the $7$ boys in a circle. The number of ways to arrange $7$ boys in a circle is $(7-1)! = 6!$.
There are $7$ gaps created between the boys in the circular arrangement.
We need to place $5$ girls in these $7$ gaps such that no two girls sit together. The number of ways to choose $5$ gaps out of $7$ is $^7C_5$.
The number of ways to arrange $5$ girls in the chosen $5$ gaps is $5!$.
Total number of ways = $6! \times ^7C_5 \times 5!$.
$= 720 \times 21 \times 120$.
$= 720 \times 21 \times 120 = 1,814,400$.
Calculating the expression in option $A$: $126 \times (120)^2 = 126 \times 14,400 = 1,814,400$.
Thus,the correct option is $A$.

(A) Let the number of ways to color $n$ vertices of a cycle graph with $k$ colors such that no two adjacent vertices have the same color be $P(n, k)$.
This is given by the formula $P(n, k) = (k-1)^n + (-1)^n(k-1)$.
Here,$n = 5$ (number of persons) and $k = 3$ (number of colors).
Substituting the values,we get:
$P(5, 3) = (3-1)^5 + (-1)^5(3-1)$
$P(5, 3) = 2^5 - 2$
$P(5, 3) = 32 - 2 = 30$.
Thus,the total number of ways is $30$.

(D) For $m$: The boys and girls can be arranged in a line alternately in two ways: ($B$ $G$ $B$ $G$ $B$ $G$ $B$ $G$ $B$ $G$) or ($G$ $B$ $G$ $B$ $G$ $B$ $G$ $B$ $G$ $B$). In each case,$5$ boys can be arranged in $5!$ ways and $5$ girls can be arranged in $5!$ ways. So,$m = 2 \times 5! \times 5! = 2 \times 120 \times 120 = 28800$.
For $n$: To arrange $5$ boys and $5$ girls in a circle such that no two boys are together,we first arrange the $5$ girls in a circle in $(5-1)! = 4! = 24$ ways. This creates $5$ gaps between them. The $5$ boys can be arranged in these $5$ gaps in $5! = 120$ ways. So,$n = 4! \times 5! = 24 \times 120 = 2880$.
Given $m = kn$,we have $28800 = k \times 2880$,which gives $k = 10$.

(C) The number of ways to arrange $n$ distinct objects in a circle is $(n-1)!$.
For a necklace,the clockwise and anti-clockwise arrangements are considered identical,so the number of ways to form a necklace with $n$ different pearls is $\frac{(n-1)!}{2}$.
Given $n = 8$,the number of ways is $\frac{(8-1)!}{2} = \frac{7!}{2}$.
$\frac{5040}{2} = 2520$.

(C) To arrange $6$ boys and $5$ girls at a round table such that no two girls sit together,we first arrange the $6$ boys in a circle.
The number of ways to arrange $6$ boys in a circle is $(6-1)! = 5! = 120$.
After arranging the boys,there are $6$ gaps created between them.
We need to place $5$ girls in these $6$ gaps such that no two girls are together.
The number of ways to choose $5$ gaps out of $6$ is $^6C_5 = 6$.
The number of ways to arrange the $5$ girls in the chosen gaps is $5! = 120$.
Therefore,the total number of ways is $120 \times 6 \times 120 = 86400$.

(C) Total number of ways to seat $10$ people around a round table is $(10-1)! = 9! = 362880$.
Let the $2$ special boys be $B_1, B_2$ and the special girl be $G_1$.
We want to find the number of arrangements where $B_1, B_2, G_1$ do not sit together.
Using the complement method,we subtract the cases where they sit together from the total.
Treating $(B_1, B_2, G_1)$ as one unit,we have $8$ units to arrange in a circle: $(8-1)! = 7! = 5040$.
Within the unit,$B_1, B_2, G_1$ can be arranged in $3! = 6$ ways.
Total cases where they sit together = $5040 \times 6 = 30240$.
Number of ways they do not sit together = $362880 - 30240 = 332640$.

(B) Total number of people $= 5 + 3 = 8$.
Total ways to arrange $8$ people on a round table $= (8-1)! = 7!$.
Now,consider the case where $B_1$ and $G_1$ sit together. Treat $(B_1G_1)$ as one unit.
Now we have $7$ units to arrange on a round table,which can be done in $(7-1)! = 6!$ ways.
Within the unit,$B_1$ and $G_1$ can be arranged in $2! = 2$ ways.
So,the number of ways they sit together $= 6! \times 2$.
The number of ways they never sit adjacent $= 7! - (2 \times 6!) = 7 \times 6! - 2 \times 6! = (7-2) \times 6! = 5 \times 6!$.

(D) Step $1$: Select $12$ friends out of $21$ to sit at the first table. The number of ways is $\binom{21}{12}$.
Step $2$: The number of ways to arrange $12$ friends at a round table is $(12-1)! = 11!$.
Step $3$: The remaining $9$ friends are to be seated at the second round table. The number of ways to arrange them is $(9-1)! = 8!$.
Step $4$: The total number of ways is $\binom{21}{12} \times 11! \times 8! = \frac{21!}{12!9!} \times 11! \times 8! = \frac{21!}{12 \times 9} = \frac{21!}{108}$.

(A) The family consists of $10$ people: $1$ mother,$1$ father,$4$ boys,and $4$ girls.
Total males = $1$ (father) + $4$ (boys) = $5$.
Total females = $1$ (mother) + $4$ (girls) = $5$.
Since the males and females must alternate,we arrange the $5$ males around the circular table in $(5-1)! = 4! = 24$ ways.
This creates $5$ spaces between the males.
We need to place the $5$ females in these $5$ spaces.
However,the mother and father must sit together.
Let the father be at position $F_1$. The mother must be in one of the two spaces adjacent to the father.
There are $2$ choices for the mother's position relative to the father.
The remaining $4$ females can be arranged in the remaining $4$ spaces in $4! = 24$ ways.
Total ways = $24 \times 2 \times 24 = 576$.

(A) Let the colours be $R, B, G$. The number of ways to colour the vertices of a cycle graph $C_n$ with $k$ colours such that no two adjacent vertices have the same colour is given by the chromatic polynomial $P(C_n, k) = (k-1)^n + (-1)^n(k-1)$.
Here,$n = 5$ (number of persons) and $k = 3$ (number of colours).
Substituting the values:
$P(C_5, 3) = (3-1)^5 + (-1)^5(3-1)$
$= 2^5 - 2$
$= 32 - 2$
$= 30$
Thus,there are $30$ ways to distribute the hats.

(A) Let the number of ways to colour a circle of $n$ vertices with $k$ colours such that no two adjacent vertices have the same colour be $P_n(k)$.
For a circular arrangement,the formula is given by $P_n(k) = (k-1)^n + (-1)^n(k-1)$.
Here,$n = 5$ (number of persons) and $k = 3$ (number of colours).
Substituting the values,we get $P_5(3) = (3-1)^5 + (-1)^5(3-1)$.
$P_5(3) = 2^5 - 1(2) = 32 - 2 = 30$.
Therefore,the total number of ways is $30$.

(A) To arrange $20$ delegates around a round table such that $2$ particular delegates sit together,we treat the $2$ delegates as a single unit.
Now,we have $18$ other delegates plus $1$ unit (the pair),making a total of $19$ units.
The number of ways to arrange $19$ units in a circle is $(19 - 1)! = 18!$.
The $2$ delegates within their unit can be arranged in $2!$ ways.
Therefore,the total number of arrangements is $2! \times 18! = 2 \times 18!$.

(A) First,arrange the $6$ men around a round table. The number of ways to arrange $n$ objects in a circle is $(n-1)!$. So,$6$ men can be seated in $(6-1)! = 5!$ ways.
There are $6$ gaps created between these $6$ men. We need to place $5$ women in these $6$ gaps such that no two women sit together.
The number of ways to arrange $5$ women in $6$ gaps is given by $P(6, 5) = \frac{6!}{(6-5)!} = 6!$.
Therefore,the total number of ways is $5! \times 6!$.

(A) The number of ways to arrange $n$ distinct objects in a circle is $(n-1)!$.
Since a garland can be flipped over,the clockwise and counter-clockwise arrangements are considered identical.
Therefore,the number of different garlands that can be prepared using $n$ different coloured flowers is given by the formula $\frac{(n-1)!}{2}$.
For $n = 5$,the number of garlands is $\frac{(5-1)!}{2} = \frac{4!}{2} = \frac{24}{2} = 12$.

(B) The number of ways to select $5$ persons out of $8$ is ${}^8C_5$.
Now,the number of ways to arrange $5$ guests at a round table is $(5-1)! = 4!$.
The number of ways to arrange the remaining $3$ guests at another round table is $(3-1)! = 2!$.
Thus,the total number of ways is ${}^8C_5 \times 4! \times 2!$.
Calculation: ${}^8C_5 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
Total ways $= 56 \times 24 \times 2 = 56 \times 48 = 2688$.
Hence,option $(B)$ is correct.

(D) Total number of people = $3 + 8 = 11$.
Number of ways to arrange $11$ people around a circle = $(11 - 1)! = 10!$.
Now,consider the case where all $3$ sisters sit together. Treat the $3$ sisters as $1$ unit.
Total units to arrange = $8$ brothers + $1$ unit of sisters = $9$ units.
Number of ways to arrange $9$ units around a circle = $(9 - 1)! = 8!$.
The $3$ sisters can be arranged among themselves in $3! = 6$ ways.
So,the number of ways where all $3$ sisters sit together = $8! \times 6$.
The number of ways where all $3$ sisters are not seated together = (Total arrangements) - (Arrangements where all $3$ sisters sit together) = $10! - (8! \times 6)$.
$10! - 6 \times 8! = (10 \times 9 \times 8!) - (6 \times 8!) = (90 - 6) \times 8! = 84 \times 8!$.