Question

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

Answer 1

\[\binom{29+5-1}{5-1}\]

Answer 2

we are discussing permutation and combination to solve these problems

Answer 3

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

Answer 4

Right, \(\dbinom {33}4={}_{33}C_4\), I'm using the binomial coefficient notation.

Recall that
\[{}_{n}C_k=\frac{n!}{k!(n-k)!}\]

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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