# Need these within 1 hour !!!

• Oct 2nd 2008, 09:15 AM
Faisal2007
Need these within 1 hour !!!
1. In how many way could 15 different books be divided equally among 3 people?
Answer in the book = 756756

2. The camera club has 5 members, and the mathematics club has 8. There is only one member common to both clubs. In how many ways could a committee of 4 people be formed with at least one member from each club?
Answer in the book = 459

3. There are 12 questions on an examination , and each student must answer 8 questions including at least 4 of the first 5 questions. How many different combinations of questions could a student choose to answer?
Answer in the book = 210
• Oct 2nd 2008, 04:03 PM
MakeANote
Hey there faisal,

Because of time zones, I'm not sure if I'm too late for you...

1. In how many way could 15 different books be divided equally among 3 people?
Answer in the book = 756756

15 books can be divided 15! ways = 1307674368000

Even though the books are different, the order within the group is not important. ie. ABCDE = ADECB (both groups contain the same letters). Each group of 5 books can be divided in 5! ways.

Hence, 15 different books are divided eually among 3 people

$\frac {15!}{5!*5!*5!} = \frac {15!}{5!^3} = 756756$ ways.

2. The camera club has 5 members, and the mathematics club has 8. There is only one member common to both clubs. In how many ways could a committee of 4 people be formed with at least one member from each club?
Answer in the book = 459

If you know what a venn diagram is, it might be helpful to draw problems like this using one...

There are 12 members all together (5 + 8 - 1(the common member) = 12)

These 12 members can be arranged in $\frac {12!}{4!} = 495$ ways.

This number represents the total permutations for the group. Taking out the other groups which contain only members from one group:

$495 - \frac {4!}{4!} - \frac {7*6*5*4}{4!} = 459$ ways.

I think you've mixed up some information for your 3rd question :)

Sorry about the delay - I'm on child-duties at home :P

Trust that helps...
• Oct 6th 2008, 03:18 PM
Faisal2007
Fixed question number 3.

Although the due date is over and i still havent gotten question # 2 and 3 i am DIEING to know the concept behind them.

• Oct 6th 2008, 03:34 PM
Henderson
On #3:

If a student does all of the first 5 problems, then he must find 3 more problems to do out of the 7 remaining problems available. There are $\frac{7*6*5}{3!} = 35$ ways to do this.

If a student does only 4 of the first five, There are $5$ ways to select those four problems (think of it as choosing a problem to skip). After those 4, he must find 4 more problems to do out of the 7 remaining problems available. There are $\frac{7*6*5*4}{4!} = 35$ ways to do this. Multiply the $5$ ways to select the first four problems by the $35$ ways to select the last four problems to get $175$.

$175$ ways to do just 4 of the first 5
$+35$ ways to do all of the first 5

$210$ ways to do 8 of the 12 problems