1. ## Probability

I'm having a lot of trouble with permutations and combinations, so I was hoping someone here could help me out with a few questions. Please make sure that you explain your working; I really want to understand this topic... The first question asks;

A queue has 4 boys and 4 girls standing in line. Find how mant different arrangements are possible if;

a) The boys and girls alternate.

b) 2 particular girls wish to stand together.

c) All the boys stand together.

d) Also find the probability that 3 particular people will be in the queue together if the queue forms randomly.

EDIT: If it's any help, here are the answers...

a) 1152
b) 10080
c) 2880
d) 3/28

EDIT: I've got a new problem now;

A table has 4 boys and 4 girls sitting around it.

a) Find the number of ways of sitting possible if the boys and girls can sit anywhere around the table.

b) If the seating is arranged at random, find the probability that;
i) 2 particular girls will sit together.
ii) All the boys will sit together.

2. Hello, Flay!

We have to "talk" our way through these problems . . .

A queue has 4 boys and 4 girls standing in line.
Find how mant different arrangements are possible if;

a) The boys and girls alternate.
There are 2 possible arrangements: .$\displaystyle BGBGBGBG\,\text{ and }\,GBGBGBGB.$

The four boys can be placed in 4! ways.
The four girls can be placed in 4! ways.

Therefore, there are: .$\displaystyle 2 \times 4! \times 4! \;=\;\boxed{1152}$ ways.

b) 2 particular girls wish to stand together.
Suppose the two girls are $\displaystyle X$ and $\displaystyle Y.$

Duct-tape them together.
Note that there are 2 possible orders: .$\displaystyle XY\,\text{ or }\,YX.$

Now we have seven "people" to arrange: .$\displaystyle \boxed{XY}\,,G,G,B,B,B,B$
. . and they can be arranged in $\displaystyle {\color{blue}7!}$ ways.

Therefore, there are: .$\displaystyle 2 \times 71\;=\;\boxed{10,080}$ ways.

c) All the boys stand together.
Duct-tape the four boys together.
. . Note that there are 4! possible orders.

Now we have five "people" to arrange.
. . There are $\displaystyle 5!$ ways.

Therefore, there are: .$\displaystyle 4! \times 5! \:=\:\boxed{2880}$ ways.

d) Find the probability that 3 particular people will be together
if the queue forms randomly.
First of all, there are 8! possible arrangements.

Suppose the three people are $\displaystyle X, Y\text{ and }Z.$
Duct-tape them together: .$\displaystyle \boxed{XYZ}$
. . Note that there are 3! orderings.

Now we have six "people" to arrange,
. . and there are 6! ways.

Hence, there are: .$\displaystyle {\color{red}3!\times6!}$ ways for $\displaystyle X,Y,Z$ to be together.

Therefore, the probability is: .$\displaystyle \frac{3!\cdot6!}{8!} \;=\;\boxed{\frac{3}{28}}$

3. Thanks very much. Now I have another problem;

A school committee is to be made up of 5 teachers, 4 students and 3 parents.

a) If 12 teachers, 25 students and 7 parents apply to be on the committee, which is chosen at random, how many possible committees could be formed?

This part I've figured out. The answer is $\displaystyle 350 658 000$.

b) If Jan and her mother both apply, find the probability that both will be chosen for the committee.

This part I'm not sure how to do. The answer is apparently $\displaystyle \frac{3036}{44275}$.

4. I figured out the last question. As it turns out, $\displaystyle \frac{3036}{44275}$ simplifies to $\displaystyle \frac{12}{175}$, which was the answer I was getting.

Now I have a new problem;

A sample of 3 coins is taken at random from a bag containing 8 ten cent coins and 8 twenty cent coins. Find the probability that a particular ten cent coin will be chosen, if 1 twenty cent and 2 ten cent coins are chosen.