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

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 $(1+1)^n$.