A probability problem by Sabhrant Sachan

$\text {According to the Question : The Sum of Outcomes must be six } \\ \text{The Minimum outcome is 1 and Max. Outcome should be 3 [For condition to be valid] } \\ \text {Using the Help of Generating functions : } \\ \text{We need to find the co-efficient of } x^6 \text{ in } (x^1+x^2+x^3)^4 \\ \implies x^4\dfrac{(1-x^3)^4}{(1-x)^4} \\ \implies \text{co-efficient of } x^2 \text { in } (1-x^3)^4(1-x)^{-4} \\ \implies \text{co-efficient of } x^2 \text { in } (1-4x^3+6x^6-4x^9+x^{12})(1+4x+10x^2+20x^3...\infty) \\ \text {Co-efficient of } x^2 \text { is } \color{#3D99F6}{\boxed{10}}$

let the numbers be p , q , r , s. p , q , r , s>=1 p + q + r +s =6 introduce a = p-1,b=q-1,c=r-1,d=s-1 then , a+b+c+d=2 no. of solutions = (2+4-1)!/(4-1)!=5!/3!=10