Find how many solutions there are to the given equation that satisfy the given condition.
y1 +y2+y3+y4=27, each yi is a nonnegative integer.

Answers

Answer 1

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


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Answers

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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

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brainly.com/question/1143565

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Answer~

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Answer:

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

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