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3 years ago

Solve the problem.

Mathematics

1 answer:

Lorico [155]3 years ago

5 0

Answer:

Step-by-step explanation:

You must use a combination:

$_nC_k=\dfrac{n!}{k!(n-k)!}$

We have n = 16, k = 3.

Substitute:

$_{16}C_3=\dfrac{16!}{3!(16-3)!}=\dfrac{13!\cdot14\cdot15\cdot16}{2\cdot3\cdot13!}\qquad\text{cancel}\ 13!\\\\=\dfrac{14\cdot15\cdot16}{2\cdot3}=\dfrac{7\cdot5\cdot16}{1}=560$

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Answer:

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Please help me on this!! 5n+17≥n–3≥–15+5n

DENIUS [597]

Answer:

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Step-by-step explanation:

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