Question

Find how many solutions there are to the given equation that satisfy the given condition.

Answer 1

Answer:

17550 solutions

Step-by-step explanation:

Given that:

y1 +y2+y3+y4=27

where;

(yi  ≥ 0 and yi $\epsilon$ ${\displaystyle \mathbb {Z} }$  )

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

$\bigg(^{27}_{4} \bigg)$

∴

The number of nonnegative integer solution is $\bigg(^{27}_{4} \bigg)$

$= \dfrac{27!}{4!(27-4)!}$

$= \dfrac{27!}{4!(23)!}$

$= \dfrac{27\times 26\times 25\times 24 \times 23 !}{4\times 3\times 2\times 1(23)!}$

$= \dfrac{27\times 26\times 25\times 24 }{4\times 3\times 2\times 1}$

$= \dfrac{421200}{24}$

= 17550 solutions