Restricted Sum

Probability Level 3

$x,y$ and $z$ are integers between $0, 100$ , inclusive.

What are all the possible non-distinct triples $(x,y,z)$ such that $x+y+z=200$

Extension : Find the number of distinct/ordered triples that satisfy the above conditions.

The answer is 5151.

2 solutions

Margaret Yu
Jun 6, 2016

$(1+x^{1}+x^{2}+x^{3}+...+x^{100})^3$

Finding the coefficient of $x^{200}$

$(1+x^{1}+x^{2}+x^{3}+...+x^{100})^3 \times (1-x)^{100} \times \frac{1}{(1-x)^{100}}$

$(1-x^{101})^3 \times (n+2)Cn \times x^{n}$

$(1 -3x^{101} + 3x^{202} - x^{303}) \times (n+2)Cn \times x^{n}$

Disregarding the last two terms since the exponent of x is > 200.

$(n+2)Cn \times x^{n} - 3 \times (n+2)Cn \times x^{101}$

For the first term n= 200, and for the second term n=99

$202C200 - 3 \times 101C99 = \boxed{5151}$

I have no idea why u did all that math can you please explain the process behind what you did? Thank you! Also why can't I use the stars and bars theorem to solve this?

Ashish Sacheti - 4 years, 12 months ago

Shalini Bhatia
Sep 14, 2017

We have x,y,z should be less than or equal to 100 So, we can write x,y,z as 100-a,100-b,100-c where a,b,c vary from 0to 100 So 100-a+100-b+100-c=200 So a+b+c=100 No. Of solutions=102C2= 5151

Restricted Sum

The answer is 5151.

2 solutions

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