Inequalities + Combinatorics = Combinequalities

Let z be any non-negative integer, such that $\displaystyle x + y + z = 100$ . The number of solutions then using "Stars and Bars" funda equals $\dbinom{102}{2}= 5151.$

Consider an inequality like:

$\Large\displaystyle\sum_{i=1}^{k-1} a_i \leq n$

We can transform this to equality by adding a non-negative integer $a_{k}$ to the $L.H.S$ like this:

$\Large\displaystyle\sum_{i=1}^k a_i = n$

By the technique of stars and bars we have total solutions for $a_i \ \forall i \ , \ 1\leq i \leq k$ given by : $\Large {{n+k-1\choose k-1}}$

Here , $k=3 \ , \ n=100$ , hence substituting the values we have :

$\Large {3+100-1\choose 3-1} = {102\choose 2} = 5151$

Well, we can calculate this using just one formula(just like rajen sir),

According to the inequality, we need to express all numbers $<100$ as the sum of two numbers, so consider this,

$0=0+0$

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$1=1+0 \\ 1=0+1$

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$2=0+2\\ 2=2+0 \\ 2=1+1$

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$3=0+3 \\ 3=3+0 \\ 3=1+2 \\ 3= 2+1 \\ . . . . . .$

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We find that each of the number can be expressed as $1$ more than the number, so we can generalize this to $n^{th}$ number.

Any number $n$ can be expressed in $n+1$ ways.

So, the inequality can be expressed as $(0+1)+(1+1)+(2+1)+(3+1). . . . . .(n+1)$ , as the last number in this question is $100$ , the number of ordered pairs is:

$1+2+3+4+5+. . . . 101 \\ = \dfrac{101×102}{2} \\=\boxed{5151}$

Let $x=0$ . We can see that $y=0,1,2,...,100$ , which is $101$ possible solutions.

Let $x=1$ . We can see that $y=0,1,2,...,99$ , which is $100$ possible solutions.

Repeating for values of $x$ up to 100, we see the total combinations comes to:

$\large {\sum _{ r=1 }^{ 101 }{ r } \\ =\frac { 1 }{ 2 } n\left( n+1 \right) \\ =\frac { 1 }{ 2 } \left( 101 \right) \left( 102 \right) \\ = \color{#20A900}{\boxed { 5151 }}}$

Put x=0 y can be anything from 0 to 100 i.e 101 possibilities for y .Then increase x to 1.Now y can go from 1 to 100 i.e 100 possibilities.Looking at the pattern the answer is the sum of natural numbers from 1 to 101 which gives 5151.