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 in

Question

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.

in progress 0
Kim Chi 5 months 2021-08-23T10:40:52+00:00 1 Answers 12 views 0

Answers ( )

  1. sori
    0
    2021-08-23T10:41:54+00:00
    Reply

    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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )