# Combination problem

1. Dec 28, 2003

### fffbone

Can anyone explain to me why k!/(k!*(k-k)!)+(k+1)!/(k!*(k+1-k)!)+(k+2)!/(k!*(k+2-k)!)+...+(n-1)!/(k!*(n-1-k)!)=n!/((k+1)!*(n-k-1)!) please. Thanks a lot!

2. Dec 29, 2003

### Hurkyl

Staff Emeritus
Look at the case where n = k+1... then the case where n = k+2...

3. Dec 29, 2003

### jcm15

logarithm problem help

can you help me with the four following problems by showing me the right procedures of doing it even though it's so troublesome? thanks alot and i would happily accept any recommended good sites from you guys for this topic.

1)4(2^2x)=8(2^x)-4
2)8(2^2x)-10(2^2x)+2
3)3*2^2x-18(2^x)+24=0
4)9^x-4(3^x)+3=0

4. Dec 29, 2003

### hedlund

1)4(2^2x)=8(2^x)-4
2)8(2^2x)-10(2^2x)+2
3)3*2^2x-18(2^x)+24=0
4)9^x-4(3^x)+3=0

1) Substitute 2^x with t and the solve the quadratic equation

2) Substitute 2^x with t and then solve the quadratic equation

3) Substitute 2^x with t and the solve the quadratic equation

4) Substitute 3^x with t and the solve the quadratic equation

5. Dec 29, 2003

### fffbone

Hurkyl,

Sorry, but I didn't quite get where you are going with n=k+1, etc. Could you please explain in more detail?

6. Dec 29, 2003

### NateTG

How about this:

Assume that $$n-k > 1$$ and simplify:
$$\frac{n!}{(k+1)!(n-k-1)!}-\frac{(n-1)!}{(k+1)!((n-1)-k-1)!}$$

Then compare it to the terms in your series.

7. Dec 29, 2003

### fffbone

Now I see, thanks.

8. Dec 29, 2003

### fffbone

I have just one more question:

Why does n!/(0!*(n-0)!)+n!/(1!*(n-1)!)+...+n!/(n!*(n-n)!)=2^n ?

9. Dec 30, 2003

### Hurkyl

Staff Emeritus
Use the binomial theorem on $(1+1)^n$.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?