Question

How many solutions are there to the equation x+y+z=17 ?They are non-negative integers  

A) 120 B)171 C)180 D)121

Answer 1

No. of solutions to the equation x+y+z=17 is given by formula

C(n+r-1, n-1)=C(3+17-1, 3-1)=C(19,2)=171

Ans (B)

Comments

There is no need to remember any formula, we can do this in this way!!

Let us take an example; x + y = 10 and we have to find out a number of  whole number solutions of it,

we can approach in this way, we have to distribute the sum of 10 to x and y in the units of 1, I mean

1111111111(sum of all these units=10) now we have to put one partition among these units because we have to break the number into two numbers(x and y)

take some instances: 111p1111111 here x =3 and y=7

                                   11p11111111 here x=2 and y=8

                                   p1111111111 here x=0 and y=10 and so on...

So Now my problem is reduced to the number of arrangements possible with these elements- 1111111111p

Therefore, number of arrangements = 11!/(10!) =11

hence, the number of whole number solutions= 11 which there are 11 ways in which we can assign a value to x and y so that sum will be equal to 10.

Similarly, we can break this question too;

x + y + z =17

take some instances: 1111(17 one's)pp

So number of arrangements = 19!/(17!*2!)= 171

Hence there are 171 ways in which we can assign a value of x,y and z so that sum will be equal to 17

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That’s a neatly explained Star Bars problem.

[Stanford Notes Section 1.2.4 | This is (From “Probability & Statistics with Applications to Computing” by Alex Tsun) | Another Open Textbook]

 

It is applicable in a  lot of scenarios such as :

1. Combinations with repetitions
2. Integer solutions of the equations
3. Multisets
4. Non (decreasing/increasing) sequences
5. Integer decompositions

etc…

All of these can correspond to the problem of distributing indistinguishable objects to distinguishable boxes.

Answer 2

$17$ identical balls (each $1$) to be put into $3$ distinct bins $(3$ variables $x,y,z).$

Solution given by $\frac{(17+3-1)!}{17!(3-1)!} = \frac{18.19}{2} = 171$

Answer 3

With Generating Function

$\left [ {\color{DarkOrange} {x^{17}}} \right ]\left ( \frac{1}{1-x} \right )\left ( \frac{1}{1-x} \right )\left ( \frac{1}{1-x} \right )$

$=\left [ {\color{DarkOrange} {x^{17}}} \right ]\left ( \frac{1}{1-x} \right )^{3}$

$=\left [ {\color{DarkOrange} {x^{17}}} \right ]\left ({1-x} \right )^{-3}$

$=1+3x+\frac{3.4}{2!}x^{2}+\frac{3.4.5}{3!}x^{3}+.............$

So Coefficient of $\left [ {\color{DarkOrange} {x^{17}}} \right ]$ is $=\frac{3.4.5.......19}{17!}=171$

Comments

Can you just explain the above approach?

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Shivateja MST

$\Large \frac{1}{(1-x)^n} = \sum_{k = 0}^{\infty} \text{C(n+k-1, k)} x^k$

so for $x^{17}$ coefficient will be $\text{C(3+17-1, 17)} = \text{C(19,17)} = 171$

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That I understood bro.My doubt is how this approach is correlated to finding no of integral solutions.

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The formula in selected answer is direct formula from the sequence we are getting from this power series.

Answer 4

I don't know if there is any direct formula for such question, but it can be solved as follows:

S1 = { (1,1, 15), (1,2,14), (1,3,13),.., (1,15,1)}     S2={(2,1,14), (2,2,13), ..(2,14,1)}        S3={(3,1,13),.. (3,13,1)}, .., S15 ={(15,1,1,)}

so, we have 1 in units place (s1) 15 times, 2 -> 14 times , 3 -> 13 times and ... to 15 ->1 time.

so answer will be -> (15+14+13+...+1)   for all possible places of digits to be covered.

so it is (15*16)/2 = 120.

 

So answer is 120.

Comments

the solution for thenon negatine innteger values of eqn of the form...

 

x1+x2+x3......+xk=(n+k-1,k-1) (permutation and combination solution)

hence

n+k-1=19

k-1=2

!19/!17*!2=171

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To count the no soultions , we need to select 17 items with x1,x2 and x3. So No of sol is equal to  17 combinations of x1,x2 and x3 elements.

c(3+17-1,17)=c(19,17)=171.

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@King Please consider zero as well. then answer will be 171