# Perms and Combs Question Help (1 Viewer)

Thank you

#### jimmysmith560

##### Le Phénix Trilingue
Moderator
Would the following working help for Question 10 and Question 11 part (a)?

Question 10:

As the tables are distinguishable, all we need to do is to assign people to each table.

Choose 5 out of 10 to be in the first table: $\bg_white \binom{10}{5}$.

Choose 5 out of the remaining 5 to be in the second table: $\bg_white \binom{5}{5}$.

After we have chosen who is to be on what table, we must arrange them. There are 4! ways of arranging those on the oak table. There are 4! ways again for the mahogany table.

$\bg_white \therefore \text{Answer}=\binom{10}{5}\times \left(4!\right)^2=145152$

Question 11:

Part (a):

Assuming the group of 4 rowers as 1 person, group of 3 basketballers as 1 person and the cricketers as 2 separate persons (So, a total of 4 persons)

Number of ways to arrange n people in a round table = $\bg_white (n-1)!$

Number of ways to arrange the groups = $\bg_white (4-1)! = 6 \:\text{ways}$.

Excluding the number of ways in which the rowers and basketballers will sit together. Assuming basketballers and rowers as one person there are $\bg_white (3-1)!\times2! = 4 \:\text{ways where they will sit with each other.}$

Hence the number of ways to arrange the groups where rowers and basketballers sit together is 2.

Number of ways to arrange basketballers among their group = $\bg_white 3! = 6$

Number of ways to arrange rowers among their group = $\bg_white 4! = 24$

Hence, total number of ways is $\bg_white 2\times6\times24 = 288$.

#### Clueless0182

##### Member
Would the following working help for Question 10 and Question 11 part (a)?

Question 10:

As the tables are distinguishable, all we need to do is to assign people to each table.

Choose 5 out of 10 to be in the first table: $\bg_white \binom{10}{5}$.

Choose 5 out of the remaining 5 to be in the second table: $\bg_white \binom{5}{5}$.

After we have chosen who is to be on what table, we must arrange them. There are 4! ways of arranging those on the oak table. There are 4! ways again for the mahogany table.

$\bg_white \therefore \text{Answer}=\binom{10}{5}\times \left(4!\right)^2=145152$

Question 11:

Part (a):

Assuming the group of 4 rowers as 1 person, group of 3 basketballers as 1 person and the cricketers as 2 separate persons (So, a total of 4 persons)

Number of ways to arrange n people in a round table = $\bg_white (n-1)!$

Number of ways to arrange the groups = $\bg_white (4-1)! = 6 \:\text{ways}$.

Excluding the number of ways in which the rowers and basketballers will sit together. Assuming basketballers and rowers as one person there are $\bg_white (3-1)!\times2! = 4 \:\text{ways where they will sit with each other.}$

Hence the number of ways to arrange the groups where rowers and basketballers sit together is 2.

Number of ways to arrange basketballers among their group = $\bg_white 3! = 6$

Number of ways to arrange rowers among their group = $\bg_white 4! = 24$

Hence, total number of ways is $\bg_white 2\times6\times24 = 288$.

Q10 makes perfect sense to me.

Regarding Q11,

Excluding the number of ways in which the rowers and basketballers will sit together. Assuming basketballers and rowers as one person there are
$image=https://latex.codecogs.com/png.latex?\bg_white+(3-1)!\times2!+=+4+\:\text{ways+where+they+will+sit+with+each+other.}&hash=7f409cb0eb4dee3b5e6cb019b2c54c92$

Hence the number of ways to arrange the groups where rowers and basketballers sit together is 2.
I dont really undertand this part. I get how you got 4 but don't understand why. Also if you got 4 why did you then write 2?

#### jimmysmith560

##### Le Phénix Trilingue
Moderator

Q10 makes perfect sense to me.

Regarding Q11,

I dont really undertand this part. I get how you got 4 but don't understand why. Also if you got 4 why did you then write 2?
No worries!

I believe it is because the 2 refers to the ways through which the groups can be arranged, whereas the 4 refers to the ways of sitting with each other.

Another way of looking at this question would be that the four rowers can sit in 4! different orderings within themselves and the basketballers can sit in 3! different orderings, while the cricketers can swap around, leading to $\bg_white 4!\times3!\times2!=288$, but perhaps someone else could explain the approach to this question better than me.