# Thread: Need these within 1 hour !!!

1. ## 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

2. 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

$\displaystyle \frac {15!}{5!*5!*5!} = \frac {15!}{5!^3} = 756756$ 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 $\displaystyle \frac {12!}{4!} = 495$ ways.

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

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

3. 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.

4. 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 $\displaystyle \frac{7*6*5}{3!} = 35$ ways to do this.

If a student does only 4 of the first five, There are $\displaystyle 5$ 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 $\displaystyle \frac{7*6*5*4}{4!} = 35$ ways to do this. Multiply the $\displaystyle 5$ ways to select the first four problems by the $\displaystyle 35$ ways to select the last four problems to get $\displaystyle 175$.

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

$\displaystyle 210$ ways to do 8 of the 12 problems

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# the camera club has 5 members

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