## anonymous one year ago How many solutions (using only nonnegative integers) are there to the following equation? x1 + x2 + x3 + x4 + x5 = 29

1. anonymous

$\binom{29+5-1}{5-1}$

2. anonymous

we are discussing permutation and combination to solve these problems

3. anonymous

I came up with this C (33)(4) but what do I do with it?"

4. anonymous

Right, $$\dbinom {33}4={}_{33}C_4$$, I'm using the binomial coefficient notation. Recall that ${}_{n}C_k=\frac{n!}{k!(n-k)!}$

5. anonymous

the thing is that I have the correct numbers but I do not know how to apply it.

6. anonymous

All you need to do is compute. Are you maybe confused as to how we arrive at $${}_{33}C_4$$?

7. anonymous

I am, how do you compute 33!??? 33x32x31x30?

8. anonymous

$\frac{33!}{4!(33-4)!}=\frac{33!}{4!29!}=\frac{33\times32\times31\times30}{4\times3\times2\times1}=11\times4\times31\times30=\cdots$

9. anonymous

ok I have $\frac{ 982,080 }{24(29!) }$ where do you get 11 x 4 x 31 x 30 this is so frustrating ok let me calculate 29! so it would be 29 x 28 x 27 x 26 = 570,024 so the answer would be ....i cant get it

10. anonymous

Thank you so much!! how did you get the 11 x 4 x 31 x 30 ? that's the correct answer 40,920

11. anonymous

$\frac{33\times32\times31\times30}{4\times3\times2\times1}=\frac{(\cancel3\times11)\times(\color{red}{\cancel8}\times4)\times31\times30}{\color{red}{\cancel4}\times\cancel3\times\color{red}{\cancel2}\times1}=11\times4\times31\times30$

12. anonymous

thank you so much. I would have never got the right answer without your help.

13. anonymous

thank you so much. I would have never got the right answer without your help.