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Leokris [45]
3 years ago
12

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

Mathematics
1 answer:
den301095 [7]3 years ago
7 0

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

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