A group of 6 boys and 3 girls sit at a round table of 9 seats for a meal.
(i) Find the probability where between any 2 girls, there is exactly 2 boys separating them.
(ii) The seats are now numbered from 1 to 9. Find the probability where the 3 girls must sit together and 3 particular boys are to be separated.
The suggested answer key says that (i) 0.0357 and (ii) 0.0214.
On
The total number of seating configurations is 9! = 362880.
For the first question. Let's put the girls in seats 1, 4, 7. We can permute the 3 girls in these seats in 3! = 6 ways. We can also permute the 6 boys in the remaining seats in 6! = 720 ways. So we have 6 * 720 = 4320 favorable seating configurations if the girls are in seats 1,4 7. But the girls can also be in seats 2, 5, 8, or the girls can be in seats 3, 6 9. In total we have a total of 3 * 4320 = 12960 favorable seating configurations.
The probability that between any 2 girls there are exactly 2 boys is 12960/362880 = 0.0357 (approximating)
For the second question. Let's put the girls in positions 1, 2, 3. We can permute them in 6 ways among themselves. The boys that have to be separated can be in positions 4, 6, 8, or in positions 4, 6, 9, or in positions 4, 7, 9, or in positions 5, 8, 9, and in each case we can permute them in 6 ways among themselves. Also , in each case, we can permute the rest of the boys in 6 ways among themselves. So if the girls are in positions 1, 2 3, we have 6 * 6 * 6 * 4 = 864 favorable seating configurations. But we can also put the girls in positions 2, 3, 4, or 3, 4, 5, or 4, 5 , 6, or 5, 6, 7, or 6, 7, 8, or 7, 8, 9, or 8, 9, 1, or 9, 1, 2. The total number of favorable seating configurations is 9 * 864 = 7776.
The probability that 3 girls must be together and 3 particular boys are to be separated is 7776/362880 = 0.0214 (approximating).
On
(i) The only possible configuration is GBBGBBGBB(G) where (G) represents the first girl back in the circular table. Hence, by fixing just one particular girl at a particular spot, permute the other two girls and 6 boys separately. Thus. the number of ways = $6!\times2!/8!$.
(ii) Here, I will label a block of 3 girls as $[GGG]$ since they are forced to be together. Next, since 3 particular boys (let's call them $A$) are to be separated, I will label the other 3 boys as $B$'s. The set-up would thus be:
$[GGG]-B-B-B-$ (- connects back to $[GGG]$). Here, since there are 4 slots for 3 $A$'s, then we have ${}^4C_3$ ways to choose for $A$. Next, if we fix $[GGG]$ for a circular permutation, we have $3!$ ways to permute the $B$'s, $A$'s and the $G$'s within the group $[GGG]$.
Hence, the total number of ways would be: ${}^4C_3 \times 3! \times 3! \times 3! \times 9$, where $9$ represents the "distinct" property created by numbering the chairs.
The probability would thus be ${}^4C_3 \times 3! \times 3! \times 3! \times 9/9!$
For (i), observe that the total number of distinct ways (up to rotation) to sit the boys and girls is $8!$.
Notice that there are only $2$ possibilities for the relative positions of the three girls. If they are labeled $A$, $B$ and $C$, then either $(1)$ $B$ is to the right of $A$ and $C$ is to the left of $A$; or $(2)$ $C$ is to the right of $A$ and $B$ is to the left of $A$.
To count all configurations that satisfy (i), it remains to count the number of ways to distribute the boys in between the girls. This can be done as follows: arrange the boys in an ordered line, then assign them seats in this order starting left from girl $A$. There are $6!$ ways to do this.
The probability is hence $2\cdot 6!/8! \simeq 0.0357$.
For (ii), observe that the total number of distinct ways to sit the boys and girls is $9!$. Since the seats are numbered, rotations of configurations are now distinct.
Notice that the group of girls can be arranged in $3!$ ways. Once a group is established, we can consider them as a single element to be sat at the numbered table.
We now consider the restriction on the boys. Label the particular boys $X$, $Y$ and $Z$. Like in (i), there are only $2$ possibilities for their relative positions. Choose one of these relative positions, and choose a seat number to assign to $X$; there are $9$ distinct ways to do so.
Given a relative position of $X$, $Y$ and $Z$ and an assignment of seat number to $X$, we must now fill the spaces between them with the remaining three boys and the girl group. We can do this in a similar way to how we did before, but notice that now there is one additional degree of freedom: because there are four 'elements' to assign and three gaps to fill, one gap will necessarily be assigned two of the elements.
We can arrange the four elements in an ordered line in $4!$ ways, and choose one of the three gaps to receive two of the elements in $3$ ways.
The probability is hence $3!\cdot 2 \cdot 9 \cdot 4! \cdot 3/9! \simeq 0.0214$.