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We want the ordered triples whose sum is $100$ .

100 stars, 2 bars, bars could be adjacent

(2 bars will divide in 3 groups, the 3 numbers)

According to the Stars and Bars Formula for ' $n$ ' stars and ' $k$ ' bars, you have to choose $n$ places for the stars, which is same as choosing $k$ places for bars, from the total $n+k$ places available. Simply $\binom{n+k}{k}$

We get that the required number will be

$\dbinom{100+2}{2} = \dbinom{102}{2} = \dfrac{101\times 102}{2} = 101\times 51 = \boxed{5151}$

Simply use combinatorics with repetition let $x_1 + x_2 + x_3 = 100$ since $x_1 , x_2 , x_3 , \geq 0$ , then we have $C(100+3-1, 100) = C(102,100)=102!/(100!2!)=(102\times101)/2=5151$

Let us draw $100$ stars which are for the jumps he made and let us have $2$ bars for the number of lanes.We are considering only $2$ bars because when we put these bars in between the stars we will have divided the $100$ stars in $3$ groups as desired.Now,if we have $100$ stars then we can put the bars in $99$ places.So,we have $99$ places to put the bars in and $2$ bars.Thus,we can divide the $100$ stars in $\dbinom{99}{2}$ ways.But by doing this we are not taking into account the cases which include $0$ jumps in $1$ or $2$ lanes.Let us say that in two lanes there were $0$ jumps.Thus,the cases are $A,B,C$ $=0,0,100)(0,100,0)(100,0,0).$ Now,let us say that he didn't jump in $1$ lane.Thus,the sum of the jumps into other lanes is $100.$ These cases are $(0),(1,2,3,4......98,99),(99,98,97,96......1)$ $(1,2,3,4......99,99),(0),(99,98,97,96......1)$ $(99,98,97....1),(1,2,3............99),(0).$ The total is $99*49+3+99*3$ $=\boxed{5151}.$

Use a Generating Function

Mathematica:

Coefficient[Sum[x^i, {i, 0, 100}]^3, x, 100]

Just have to find the no. of non negative integral solutions of a+b+c=100 that is (100+3-1)C(3-1)=102C2=5151