# Stats help (combinations/ order prb)

• Nov 5th 2007, 10:25 AM
Stats help (combinations/ order prb)
Due in 3 hours….. (1pm mt time)
I have played with these for an hr and i have completed all homework except thiese 2 problems please help.....

Question 1 (complete all 3 parts)
From a group of 9 novels 7 history books and 4 self help books:

A)In how many different ways can 4 books be chosen so that 2 of them are novels and 2 of them are history books?

B) In how many different ways can 5 books be chosen so that there are either 0 or 1 novels ?

C) If 5 books are chosen at random, what is the probability that at least one of each type is chosen?

Question 2 (Completed all 4 parts)
A line is formed from a group of 4 girls and 3 boys:

A) How many different lines begin with a girl and end with a boy?

B) How many different lines have the individuals alternating by gender?

C)If all lines are equally likely, what is the probability that the line begins with a girl and ends with a boy

D)If all lines are equally likely what is the probability that the line begins and ends with individuals of the same gender?
• Nov 5th 2007, 12:33 PM
Soroban

Quote:

1) From a group of 9 novels, 7 history books, and 4 self-help books:

A) In how many different ways can 4 books be chosen so that:
2 of them are novels and 2 of them are history books?

There are: . ${9\choose2} = 36$ ways to choose two Novels.
There are: . ${7\choose2} = 21$ ways to choose two History books.

Therefore, there are: . $36 \times 21 \:=\:\boxed{756\text{ ways}}$

Quote:

B) In how many different ways can 5 books be chosen so that:
there are either 0 or 1 novels ?

To have 0 Novels, we choose 5 from the other 11 books.
. . There are: . ${11\choose5} \:=\:462$ ways.

To have 1 Novel, we choose 1 Novel from the 9 Novels and 4 from the other 11 books.
. . There are: . ${9\choose1}{11\choose4} \:=\:9\cdot330 \:=\:2970$ ways.

Therefore, there are: . $462 + 2970 \:=\:\boxed{3432\text{ ways}}$

Quote:

C) If 5 books are chosen at random, what is the probability that
at least one of each type is chosen?

I'm still working on this part . . .