Question

How many solutions (using only nonnegative integers) are there to the following equation?
x1 + x2 + x3 + x4 + x5 = 31

Answer 1

I need an numerical number. Infinite doesn't work

Answer 2

what do x1 ,x2 ,....mean? \(x^2\)?

Answer 3

or \(x_2\)? just subscript?

Answer 4

https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.986317.html
I gotchu fam :3

Answer 5

I guess it's written as x

Answer 6

From what i understand its essentially
31+0+0+0+0=31
30+1+0+0+0=31
29+1+1+0+0=31 so on and so forth

Answer 7

hint: http://www.mathsisfun.com/algebra/fundamental-theorem-algebra.html

Answer 8

we need 5 numbers which add up to 31

we can also say that we need to divide 31 into 5 parts

these are 31 blocks-

\(\square \square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\square\)

we need to find number of ways in which we can divide these blocks into 5 parts.

one such way is this-

\(\square \square\square {\Huge\color{red}|}\square\square\square\square\square{\Huge\color{red}|}\square\square\square\square\square{\Huge\color{red}|}\square\square\square\square\square\square\square\square\square\square{\Huge\color{red}|}\square\square\square\square\square\square\square\square\)

the 4 red sticks are dividing the 31 blocks into 5 parts :)
the drawing above is just a case i took, we can change the position of sticks and they will always divide the 31 blocks into 5 parts

so we can say that the total number of solutions is equal to the total number of ways in which we can distribute the sticks in this system-
\(\square \square\square {\Huge\color{red}|}\square\square\square\square\square{\Huge\color{red}|}\square\square\square\square\square{\Huge\color{red}|}\square\square\square\square\square\square\square\square\square\square{\Huge\color{red}|}\square\square\square\square\square\square\square\square\)

there are 31 blocks
so therefore there are 30 spaces in between them where we can place the 4 sticks

so total number of ways must be this-> \(\Large ~^{30}C_4\)

Answer 9

\[\left(\begin{matrix}30 \\ 4\end{matrix}\right) = ? \] This is where I got but am lost past this

Answer 10

I think thats what you meant by 30 C 4

Answer 11

:D yes thats what i meant
\(\left(\begin{matrix}30 \\ 4\end{matrix}\right) \) can be written like this-

\(\left(\begin{matrix}30 \\ 4\end{matrix}\right) = \large \frac{\lfloor{30} \rfloor}{\lfloor{4} \rfloor \lfloor{30-4} \rfloor}\)

Answer 12

How do I come to a solution from there?

Answer 13

do you know about factorials?

Answer 14

\(\lfloor{n}\rfloor\) denotes \(\color{blue}{n ~factorial}\)
we can also write it like this- \(\lfloor{n} \rfloor=n!\)
^this is just a different notation of factorial thingy

what "n factorial" is really->
\(n!=n \times (n-1) \times (n-2) \times ...... (3) \times (2) \times (1)\)

its just the multiplication of all numbers starting from 1 till n :)

Answer 15

all *natural numbers

Answer 16

So to solve \[30i \div 4i(30i-4i) \] or \[30i \div (4i \times 26i)\]

Answer 17

its not "\(i\)" its denoted by the exclamation mark "\(!\)"

and it can be solved like this-

\(\Large \frac{30!}{4!(30-4)!}\)

=\(\Large \frac{30!}{4!(26)!}\)

=\(\Large \frac{30 \times 29 \times ..... \times \color{red}{ 26 \times 25 \times......2 \times 1}}{4! {(\color{red}{26 \times 25 \times...... \times 2 \times 1})}}\)

the red part from numerator and denominator gets cancelled and you get this-

=\(\Large \frac{30 \times 29 \times 28 \times 27}{ 4 \times 3 \times 2 \times 1}\)

now from here you must simplify this and get the answer =]

Answer 18

\[657720\div24=27405\] When I put that in it shows incorrect

Answer 19

this calculation seems length tho xP
you can use a calculator if its allowed \(\Huge \ddot \smile\)

Answer 20

sorry I dont know I kept putting i instead of !

Answer 21

Am I doing the math incorrectly or missing a step?

Answer 22

no
you did it correct
i made a mistake in this part- http://prntscr.com/ahty2q

it must be this-
there are 31 blocks and 4 sticks
so total number of ways to distribute 31 into 5 parts must be equal to total number of combinations of these 31 blocks and 4 sticks

so
total number of combinations of these 31 blocks and 4 sticks = \(\large ~^{35}C_4\)

Answer 23

So does that just change the equation to 35!/4!(30-4)!
or (35-4)!

Answer 24

it changes it to this-
\(\large \frac{35!}{4!(35-4)!}\)

Answer 25

Came to 52360. That correct?

Answer 26

35*34*33*32=1256640
1256640/24=52360

Answer 27

yeah that looks alright to me

Answer 28

Perfect thank you very much

Answer 29

:'D no prblem lol