How Can Combinatorics Solve Arrangement and Committee Selection Problems?

[rover]

hi,
I have been trying to solve the following questions (and similar) for 2 hours now without any major success.

the questions are:

1) In how many different ways can 7 different books be arranged in a row if
a. 3 specific books must be together
b. two specified books must occupy the ends

(I am TOTALY stuck! on this one!)

the other question i need help with is

2) A committee of 5 men and 4 women is to be selected from 9 men and 8 women. Amongst these is a married couple. if the married couple can not serve together, how many committees can be formed

(the total number of committee that can be formed is 9C5 * 8C4, don't know what to do next! )Thanks in advance!

Rover

BTW this not for homework, I got a test soon !

 

[rover]

could someone please answer... at least give your suggestion about how to solve the question..kind'a desperate!

 

[rover]

this what i have done
a. 4!*3! (gives wrong answer) ... OHHH

 

[eok20]

For 1.a You want to consider the three books as a single book in a sense and then multiply by how many ways you can arrange the set of three books. There are 5! ways to arrange the set of three books with the other four, not 4!.

For 1.b I think you should consider how many ways you can arrange the 5 books that are not on the ends and then mutiply by how many ways the two ends could be.

For 2 try figuring out how many arrangements have the married couple and subtract this from the total amount you found.

 

[rover]

thanks eok20!

but i am still having problems with ques. 2

 

[eok20]

So for 2, you want to subtract the number of arrangements that have the couple from the total. The number of arrangements that have the couple would be how many ways of the 8 remaining men you can choose 4 (because the married man is already there) multiplied by how many ways of the 7 remaining women you can choose 3.

 

[HallsofIvy]

1) In how many different ways can 7 different books be arranged in a row if
a. 3 specific books must be together
Imagine the 3 books as a single object. Including the other 4 books, you have 5 objects- there are 5!= 120 ways to put those in a row.
HOWEVER, there are also 3!= 6 ways to put those 3 books together.

b. two specified books must occupy the ends
Ignore the two specified books. There are 5 books left and so 5!= 120 ways to put those books in a row. Now there are also 2!= 2 ways to choose at which end to put each of the two specified books.

2) A committee of 5 men and 4 women is to be selected from 9 men and 8 women. Amongst these is a married couple. if the married couple can not serve together, how many committees can be formed

If you remove the couple, there are 8 men and 7 women left. How many ways can you form a committee of 5 men and 4 women from those? (How many commitees include neither the husband or wife?)

Now assume the husband is on the committee. Since his wife cannot be on the committee also that also leaves 8 men and 7 women. How many ways can you choose the remaining 4 men and 4 women on the committee?

Now assume that the wife is on the committee. Since her husband cannot be on the committee also that leaves 8 mean and 7 women. How many ways can you choose the remaining 5 men and 3 women on the committee?

Finally, how do you combine those 3 numbers?