Answer:
17550 solutions
Step-by-step explanation:
Given that:
y1 +y2+y3+y4=27
where;
(yi ≥ 0 and yi [tex]\epsilon[/tex] [tex]{\displaystyle \mathbb {Z} }[/tex] )
The no. of a nonnegative integer determines the number of ways to choose 27 objects from (4) distinct objects with repetition regardless of the order.
i.e
[tex]\bigg(^{27}_{4} \bigg)[/tex]
∴
The number of nonnegative integer solution is [tex]\bigg(^{27}_{4} \bigg)[/tex]
[tex]= \dfrac{27!}{4!(27-4)!}[/tex]
[tex]= \dfrac{27!}{4!(23)!}[/tex]
[tex]= \dfrac{27\times 26\times 25\times 24 \times 23 !}{4\times 3\times 2\times 1(23)!}[/tex]
[tex]= \dfrac{27\times 26\times 25\times 24 }{4\times 3\times 2\times 1}[/tex]
[tex]= \dfrac{421200}{24}[/tex]
= 17550 solutions
