Circular permutations Questions in English

(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) = $10! - (8! \times 6)$.
$10! - 6 \times 8! = (10 \times 9 \times 8!) - (6 \times 8!) = (90 - 6) \times 8! = 84 \times 8!$.

(A) First,arrange the $5$ boys around a circular table. The number of ways to arrange $n$ objects in a circle is $(n-1)!$. So,the number of ways to arrange $5$ boys is $(5-1)! = 4!$.
There are $5$ spaces created between the $5$ boys where the $4$ girls can sit to ensure no two girls sit together.
The number of ways to arrange $4$ girls in these $5$ spaces is given by $P(5, 4) = \frac{5!}{(5-4)!} = \frac{5!}{1!} = 5!$.
Therefore,the total number of ways is $4! \times 5!$.

(A) First,arrange the $6$ distinct white roses in a circle. The number of ways to arrange $n$ distinct items in a circle is $(n-1)!$. So,$6$ white roses can be arranged in $(6-1)! = 5! = 120$ ways.
There are $6$ gaps created between these $6$ white roses. We need to place $6$ distinct red roses in these $6$ gaps so that no two red roses are together. The number of ways to arrange $6$ distinct red roses in $6$ gaps is $6! = 720$ ways.
Since it is a garland,the clockwise and anti-clockwise arrangements are considered the same. Therefore,we divide the total number of linear arrangements by $2$.
The total number of ways $= \frac{5! \times 6!}{2} = \frac{120 \times 720}{2} = \frac{86400}{2} = 43200$.

(C) First,arrange the $9$ men around a circular table,which can be done in $(9-1)! = 8!$ ways.
There are $9$ gaps created between the $9$ men.
To ensure no two women sit together,we must place the $5$ women in these $9$ gaps.
The number of ways to arrange $5$ women in $9$ gaps is given by $^9 P_5$.
Therefore,the total number of ways is $8! \times ^9 P_5$.

(C) First,arrange the $2$ red and $3$ white roses in a circle. The number of ways to arrange $5$ distinct items in a circle is $(5-1)! = 4! = 24$.
There are $5$ gaps created between these $5$ roses.
We need to place $5$ yellow roses in these $5$ gaps such that no two yellow roses are together.
The number of ways to arrange $5$ distinct yellow roses in $5$ gaps is $5! = 120$.
Since it is a garland,the clockwise and anti-clockwise arrangements are considered the same,so we divide by $2$.
Total number of ways $= \frac{4! \times 5!}{2} = \frac{24 \times 120}{2} = 1440$.

(A) Fix $A$ at one position. Let the positions be $1, 2, 3, 4, 5, 6$ in clockwise order,with $A$ at position $1$. The immediate right of $A$ is position $6$ (since they face the centre).
Case $1$: $E$ is at position $6$. Then $E$ must have $F$ or $D$ at position $5$.
Subcase $1.1$: $F$ is at position $5$. The remaining $3$ persons $(B, C, D)$ can be arranged in the remaining $3$ positions in $3! = 6$ ways.
Subcase $1.2$: $D$ is at position $5$. The remaining $3$ persons $(B, C, F)$ can be arranged in the remaining $3$ positions in $3! = 6$ ways.
Total for Case $1 = 6 + 6 = 12$ ways.
Case $2$: $F$ is at position $6$. Then $E$ must have $F$ or $D$ on his immediate right. Since $F$ is at position $6$,$E$ cannot be at position $5$ (because $F$ is at $6$,not $5$). Thus,$E$ must be at some other position $k$ such that position $k-1$ (clockwise) is $F$ or $D$.
By checking the remaining positions,we find $6$ valid arrangements for Case $2$.
Total ways $= 12 + 6 = 18$.

(B) First,arrange the $6$ distinct white roses in a circle. The number of ways to arrange $n$ distinct objects in a circle is $(n-1)!$. So,the number of ways to arrange $6$ white roses is $(6-1)! = 5! = 120$.
There are $6$ gaps created between the $6$ white roses in the circle.
We need to place $5$ distinct red roses in these $6$ gaps such that no two red roses are together. The number of ways to choose $5$ gaps out of $6$ is $^6C_5 = 6$.
The number of ways to arrange $5$ distinct red roses in the chosen $5$ gaps is $5! = 120$.
Since the garland can be flipped (clockwise and anti-clockwise arrangements are considered the same),we divide by $2$.
Total number of ways $= \frac{5! \times ^6C_5 \times 5!}{2} = \frac{120 \times 6 \times 120}{2} = \frac{86400}{2} = 43200$.

(D) First,arrange the $8$ men around a circular table. The number of ways to arrange $n$ objects in a circle is $(n-1)!$. So,the number of ways to arrange $8$ men is $(8-1)! = 7!$.
After arranging the men,there are $8$ gaps created between them.
To ensure no two women sit together,we must place the $4$ women in these $8$ gaps.
The number of ways to choose and arrange $4$ women in $8$ gaps is given by the permutation formula ${}^{8}P_{4}$.
Therefore,the total number of ways is $7! \times {}^{8}P_{4}$.

(C) Total number of ways to arrange $20$ persons around a round table is $(20-1)! = 19!$.
Fix person $A$ at one position.
There are $19$ remaining seats for person $B$.
For there to be exactly $6$ persons between $A$ and $B$,$B$ must sit in a specific position relative to $A$.
Counting $6$ seats clockwise from $A$,the $7$th seat is occupied by $B$.
Counting $6$ seats counter-clockwise from $A$,the $7$th seat is also occupied by $B$.
Thus,there are $2$ favorable positions for $B$ out of $19$ possible seats.
Therefore,the probability is $\frac{2}{19}$.

(B) Total number of ways to arrange $5$ red and $5$ white roses in a garland is the number of ways to arrange $10$ distinct items in a circle,which is $(10-1)! = 9!$.
Since the roses are of different sizes,we consider them distinct.
For the roses to be placed alternately,we first fix the $5$ red roses in a circle in $(5-1)! = 4!$ ways.
This creates $5$ gaps between the red roses.
We can place the $5$ white roses in these $5$ gaps in $5!$ ways.
Thus,the number of favorable arrangements is $4! \times 5!$.
The probability is $\frac{4! \times 5!}{9!} = \frac{24 \times 120}{362880} = \frac{2880}{362880} = \frac{1}{126}$.

(D) Total number of ways to arrange $8$ teachers and $4$ students around a circular table is $(8+4-1)! = 11!$.
To ensure no two students sit together,we first arrange the $8$ teachers in a circle,which can be done in $(8-1)! = 7!$ ways.
This creates $8$ gaps between the teachers. We need to place $4$ students in these $8$ gaps,which can be done in $^8C_4$ ways.
The students can be arranged among themselves in $4!$ ways.
Thus,the number of favorable arrangements is $^8C_4 \times 4! \times 7!$.
The required probability is $\frac{^8C_4 \times 4! \times 7!}{11!} = \frac{\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \times 4! \times 7!}{11 \times 10 \times 9 \times 8 \times 7!} = \frac{70 \times 24 \times 7!}{11 \times 10 \times 9 \times 8 \times 7!} = \frac{7}{33}$.

(A) The word "$COMBINATIONS$" has $12$ letters: $C, O, M, B, I, N, A, T, I, O, N, S$.
Consonants: $C, M, B, N, N, T, S$ ($7$ letters,with $N$ repeating twice).
Vowels: $O, I, A, I, O$ ($5$ letters,with $O$ repeating twice and $I$ repeating twice).
First,arrange the $7$ consonants in a circle. The number of ways to arrange $n$ items in a circle is $(n-1)!$. Since $N$ repeats twice,the number of ways is $\frac{(7-1)!}{2!} = \frac{6!}{2!}$.
There are $7$ gaps created between these $7$ consonants. We need to place the $5$ vowels in these $7$ gaps such that no two vowels are together. The number of ways to choose $5$ gaps out of $7$ is ${ }^{7}C_{5}$.
The $5$ vowels can be arranged in these $5$ chosen gaps in $5!$ ways. Since $O$ and $I$ each repeat twice,the number of arrangements is $\frac{5!}{2!2!}$.
Total number of ways $= \frac{6!}{2!} \times { }^{7}C_{5} \times \frac{5!}{2!2!} = \frac{6!}{2!} \times \frac{7!}{5!2!} \times \frac{5!}{2!2!} = \frac{7!6!}{(2!)^4}$.

(B) First,arrange the $5$ boys around a circular table. The number of ways to arrange $n$ objects in a circle is $(n-1)!$. So,the number of ways to arrange $5$ boys is $(5-1)! = 4! = 24$.
There are $5$ spaces created between the $5$ boys where the $4$ girls can sit to ensure no two girls sit together.
The number of ways to arrange $4$ girls in these $5$ spaces is given by $P(5, 4) = \frac{5!}{(5-4)!} = \frac{120}{1} = 120$.
Therefore,the total number of ways is $4! \times 120 = 24 \times 120 = 2880$.
This is equivalent to $4! \times 5!$.

(B) To seat $5$ boys and $5$ girls around a circular table such that no two boys and no two girls are together,they must sit in an alternating fashion ($B$-$G$-$B$-$G$-$B$-$G$-$B$-$G$-$B$-$G$).
First,fix one boy at a position on the circular table. This can be done in $1$ way.
The remaining $4$ boys can be arranged in the remaining $4$ seats in $(4-1)! = 3! = 6$ ways.
There are $5$ gaps between the boys where the $5$ girls can be seated.
The $5$ girls can be arranged in these $5$ gaps in $5! = 120$ ways.
Therefore,the total number of ways is $1 \times 6 \times 120 = 720$.
Wait,re-evaluating: The number of ways to arrange $n$ boys and $n$ girls in a circle such that they alternate is $(n-1)! \times n!$.
For $n=5$,this is $(5-1)! \times 5! = 4! \times 5! = 24 \times 120 = 2880$.
Thus,the correct option is $B$.

(B) Total people $= 6 \text{ men} + 4 \text{ women} = 10 \text{ people}$.
First,calculate the total number of ways to seat $10$ people around a circular table,which is $(10-1)! = 9!$.
Next,calculate the number of ways where a particular man and a particular woman sit adjacent to each other.
Treat the particular man and woman as a single unit. Now we have $9$ units to arrange around a circular table,which can be done in $(9-1)! = 8!$ ways.
Within the unit,the man and woman can be arranged in $2! = 2$ ways.
So,the number of ways they sit together is $2 \times 8!$.
The number of ways they never sit adjacent is the total ways minus the ways they sit together:
$9! - (2 \times 8!) = (9 \times 8!) - (2 \times 8!) = (9 - 2) \times 8! = 7 \times 8!$.

(D) Total number of girls $= 3 + 8 = 11$.
Number of ways to arrange $11$ girls around a circle is $(11 - 1)! = 10!$.
To find the number of ways where the $3$ sisters are $NOT$ seated together,we use the complement method: Total arrangements $-$ Arrangements where $3$ sisters are together.
Treating the $3$ sisters as a single unit,we have $8 + 1 = 9$ units to arrange in a circle,which can be done in $(9 - 1)! = 8!$ ways.
The $3$ sisters can be arranged among themselves in $3!$ ways.
So,arrangements where $3$ sisters are together $= 8! \times 3!$.
Required number of ways $= 10! - (8! \times 3!) = 10! - (8! \times 6)$.
$= 8! \times (10 \times 9 - 6) = 8! \times (90 - 6) = 8! \times 84$.

(A) The number of circular permutations of $n$ distinct objects is $(n-1)!$.
For a necklace,the clockwise and anti-clockwise arrangements are considered identical,so the number of ways is $\frac{(n-1)!}{2}$.
Here,$n = 8$.
Number of ways = $\frac{(8-1)!}{2} = \frac{7!}{2} = \frac{5040}{2} = 2520$.