1. ## Tricky Combination Questions

1. Solve the equation for n algebraically.

nCn-2 = 10

Would I have to divide the left side by (n)(n)(n-1)(n-2), which would cancel out (n) and (n-2). Im stuck with what to do after that.

2. A pencil case can hold seven pens and eight pencils. Meg chooses either a pen or pencil. Then John is to choose a pen and a pencil. In which case (based on Andrea's initial choice) would John have to have the greater number of choices.

I think Meg's choice would be: 7C1 x 8C0 + 7C0 x 8C1
Then i get stuck on John. Would it be 6C1 x 7C1

Thanks.

2. Originally Posted by CroatBoy
2. A pencil case can hold seven pens and eight pencils. Meg chooses either a pen or pencil. Then John is to choose a pen and a pencil. In which case (based on Andrea's initial choice) would John have to have the greater number of choices.[/SIZE][/SIZE][/SIZE][/SIZE]
If Meg chooses a pen then the total number of ways is (7)(6)(8).
If Meg chooses a pencil then the total number of ways is (8)(7)(7).
No she does either and not both,so add.

3. Originally Posted by Plato
If Meg chooses a pen then the total number of ways is (7)(6)(8).
If Meg chooses a pencil then the total number of ways is (8)(7)(7).
No she does either and not both,so add.
I dont understand what mean. Why would you multiply 7x6x8 and 8x7x7?

4. Originally Posted by CroatBoy
I dont understand what mean. Why would you multiply 7x6x8 and 8x7x7?
There are 7 ways for Meg to select a pen. Once that is done the John has 6 ways to select a pen and 8 ways to pick a pencil: (7)(6)(8).

5. Originally Posted by Plato
There are 7 ways for Meg to select a pen. Once that is done the John has 6 ways to select a pen and 8 ways to pick a pencil: (7)(6)(8).
Alright. I see what you mean, thanks alot.

6. Originally Posted by CroatBoy
Alright. I see what you mean, thanks alot.
$\displaystyle \frac{n!}{(n-2)! 2!} = 10$
$\displaystyle \frac{n(n-1)}{2}=10$
$\displaystyle n^2 - n - 20 = 0$
$\displaystyle n = -4,5$
Only the positive one makes sense, so pick $\displaystyle n=5$.