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Given a,b,c,d belonging to the set of whole numbers and given the equation

$a+b+c+d=100$

How many solutions like

$(a,b,c,d)=(80,10,5,5)\; ; \; (a,b,c,d)=(0,1,2,97)$

exist? We can repeat elements and the order does not matter.

sai-kartik
  • 2,090
Duarte Arribas
  • 459
  • By “the order does not matter” do you mean $(10,20,30,40)$ and $(40,30,20,10)$ are different solutions or the same? – Thomas Andrews Jun 13 '20 at 18:10
  • they're the same – Duarte Arribas Jun 13 '20 at 18:17
  • This is a linear equation with infinitely many solutions. You can easily prove it with induction. – Jeremy Jun 13 '20 at 18:19
  • If "whole numbers" means positive integers, you are asking for the number of integer partitions of $100.$ This was asked in a more general form in https://math.stackexchange.com/questions/2429534/number-of-integer-partitions – David K Jun 13 '20 at 18:23
  • Whole numbers are natural numbers including zero. – UmbQbify Jun 13 '20 at 19:10
  • If you are including 0 then you can just find the number of integral partitions if 101 and that is equivalent to this answer – atul ganju Jun 13 '20 at 19:14
  • OK, whole numbers are non-negative integers. Actually that is what gives you all integer partitions of $100$; if you required the four terms to be positive then you would have only the integer partitions into four parts. – David K Jun 13 '20 at 19:31

1 Answers1

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We're looking for the non-negative integer solutions for the equation $x_1 + x_2 + x_3 + x_4 = 100$.

Instead of using numbers for writing the solutions, we will use strokes, so for instance we represent the solution $ x_1 = 1, x_2 = 1, x_3 = 1, x_4 =97 $, or 1 + 1 + 1 + 97, like this: | + | + | + ||| $ \cdots$ [97 strokes].

Now, each possible solution is an arrangement of 100 strokes and 3 plus signs, so the number of arrangements is P(103; 100, 3) = $\frac{103!}{100! *3!}$

The general solution for such questions -combinations with repitition- is: $\mathit{P}(n+r-1;r,n-1) = \frac{(n+r-1)!}{r!(n-1)!}=\binom{n+r-1}{r}$