# 1.)5g=__mg2.)875mg=__g3.)15kg=__g4.)25000mg=__g5.)85cg=__mg​

garcikyl000   ·   25.02.2021 15:40
01.07.2019 05:30

there is only one way to arrange all the eight $$b's$$ relative to each other because they are not distinct. we have

$$\underline{\ \ }b\underline{\ \ }b\underline{\ \ }b\underline{\ \ }b\underline{\ \ }b\underline{\ \ }b\underline{\ \ }b\underline{\ \ }b\underline{\ \ }$$

now there are nine places to place an $$a$$ (and not more than one $$a$$ in one place).   because we have four $$a$$ symbols, we need to choose four of these nine places to insert an $$a$$. order does not matter because the $$a$$ are not distinct.

thus there are 1 · c(9,4) ways, or, if we calculate it,

$$1\cdot c(9,4) = \dfrac{9! }{4! (9-4)! } = \dfrac{9\cdot8\cdot7\cdot6\cdot5! }{4! 5! } = \dfrac{9\cdot8\cdot7\cdot6}{4\cdot3\cdot2\cdot1} = 126.$$

thus there are 126 ways to form such "words."

26.06.2019 08:30

345

step-by-step explanation:

25.06.2019 19:00

step-by-step explanation:

cats

25.02.2021 15:40

Step-by-step explanation:

1.)5000mg

2.)0.875g

3.)15000g

4.)25g

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