# Thread: *URGENT* Organized Counting / Permutations / Combinations Questions help.

1. ## *URGENT* / Permutations / Combinations Questions help. Still no replies !!!

1. In how many ways can 6 boys and 4 girls be seated at a round table?

2. In how many ways can a student committee of 8 boys and 9 girls be selected from 13 boys and 15 girls?

3.A quarterback has a series of six plays possible. If the coach asks the quarterback not to repeat any plays in a game, how many different orders of plays is possible?

4. In how many ways can six girls and six boys in a choir stand in a line if boys and girls must alternate positions?

5. From a group of 27 contest winners, 18 will be chosen to receive a computer prize package. In how many ways can you choose these 18 from 27 winners?

6. How many ways can you pick a vice president, a president and a treasurer from a group of 8 boys and 7 girls if:
- There are no restrictions
-There must be at least 2 boys chosen
-There must be at least 1 girl chosen

7. In how many different ways can you arrange 9 flags in a row if there are 4 red flags and 5 blue flags?

8. If there are 14 books on the shelf, in how many ways can you arrange them if:
-With no restrictions
-They are not in alphabetical order by author (assume different authors)

9. How many different ways can you arrange ALL the letters of the word PARALLELEPIPED ?
10. In a small drama club of 18 students, 8 students are to be chosen as “Extras” in the play 4 students are to be selected as stage heads, and the rest will be on the promotions committee. In how many ways can you do this?

11. If the call letters of Canadian broadcasting stations must begins with the letter C, how many different names can be made for a station using 3 call letters where repetitions are allowed?
Again, This forum has been a great help.
Thanks to the helpers =)

2. Hello, Faisal2007!

Most of these are basic counting problems.
I must assume you are familiar with the formulas.

I'll walk you through a bunch of them . . .

1. In how many ways can 6 boys and 4 girls be seated at a round table?
There are 10 people.
(Their gender is not considered.)

The first person can sit anywhere . . . it doesn't matter.

The other 9 people can be seated in $9!$ ways.

Therefore, there are: . $9! \:=\:\boxed{362,880}$ ways.

2. In how many ways can a student committee of 8 boys and 9 girls
be selected from 13 boys and 15 girls?
Choose 8 boys from 13 . . . there are: . ${13\choose8}$ ways.

Choose 9 girls from 15 . . . there are: . ${15\choose9}$ ways.

Therefore, there are: . ${13\choose8}{15\choose9} \:=\:(12876)(5005) \:=\:\boxed{6,441,435}$ ways.

3.A quarterback has a series of six plays possible.
If the coach asks the quarterback not to repeat any plays in a game,
how many different orders of plays is possible?
Assuming he uses all six plays, there are $6! = \boxed{720}$ possible orders.

4. In how many ways can six girls and six boys in a choir stand in a line
if boys and girls must alternate positions?

There are two basic arrangements: . $BGBGBGBGBGBG\:\text{ or }\:GBGBGBGBGBGB$

In each, the boys can be seated in 6! ways
. . and the girls can be seated in 6! ways.

Therefore, there are: . $2 \cdot 6! \cdot 6! \:=\:\boxed{1,036,800}$ ways.

5. From a group of 27 contest winners, 18 will be chosen.
In how many ways can you choose these 18 from 27 winners?
Answer: . ${27\choose18}$

6. How many ways can you pick a President, Vice President and Treasurer
from a group of 8 boys and 7 girls if:

(a) There are no restrictions
(b) There must be at least 2 boys chosen
(c) There must be at least 1 girl chosen

(a) There are 15 candidates.

There are 15 choices for the President,
. . 14 choices for the Vice President,
. . 13 choices for the Treasure.

Therefore, there are: . $15 \cdot 14 \cdot 13 \:=\:\boxed{2730}$ ways.

(b) "at least 2 boys" means "2 boys or 3 boys".
If two boys (and one girl) are chosen, there are: . ${8\choose2}{7\choose1} = 196$ choices.
Since each can hold a different office, there are $3! = 6$ assignments.
. . Hence, there are: . $196 \cdot 6 \:=\:1176$ ways for two boys.

If three boys are chosen, there are: . $8 \cdot 7 \cdot 6 \:=\:336$ ways.

Therefore, there are: . $1176 + 336 \:=\:\boxed{1512}$ ways.

(c) The opposite of "at least one girl" is "no girls" (or "all boys").

From (a) we know that there are $2730$ possible outcomes.

From (b) we that there are $336$ outcomes with all boys.

Therefore, "at least one girl" has: . $2730 - 336 \:=\:\boxed{2394}$ ways.