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Alex17521 [72]
3 years ago
10

What does f equal in this equation? 6f-12=-4f+6

Mathematics
2 answers:
kakasveta [241]3 years ago
7 0
F = 1.8 is the answer
nadya68 [22]3 years ago
5 0
6f - 12 = -4f + 6...add 4f to both sides
6f + 4f - 12 = 6....add 12 to both sides
6f + 4f = 6 + 12 ...simplify
10f = 18...divide both sides by 10
f = 18/10 which reduces to 9/5 or 1 4/5 <=
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3 years ago
Suppose <img src="https://tex.z-dn.net/?f=m" id="TexFormula1" title="m" alt="m" align="absmiddle" class="latex-formula"> men and
ollegr [7]

Firstly, we'll fix the postions where the n women will be. We have n! forms to do that. So, we'll obtain a row like:

\underbrace{\underline{~~~}}_{x_2}W_2 \underbrace{\underline{~~~}}_{x_3}W_3 \underbrace{\underline{~~~}}_{x_4}... \underbrace{\underline{~~~}}_{x_n}W_n \underbrace{\underline{~~~}}_{x_{n+1}}

The n+1 spaces represented by the underline positions will receive the men of the row. Then,

x_1+x_2+x_3+...+x_{n-1}+x_n+x_{n+1}=m~~~(i)

Since there is no women sitting together, we must write that x_2,x_3,...,x_{n-1},x_n\ge1. It guarantees that there is at least one man between two consecutive women. We'll do some substitutions:

\begin{cases}x_2=x_2'+1\\x_3=x_3'+1\\...\\x_{n-1}=x_{n-1}'+1\\x_n=x_n'+1\end{cases}

The equation (i) can be rewritten as:

x_1+x_2+x_3+...+x_{n-1}+x_n+x_{n+1}=m\\\\&#10;x_1+(x_2'+1)+(x_3'+1)+...+(x_{n-1}'+1)+x_n+x_{n+1}=m\\\\&#10;x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-(n-1)\\\\&#10;x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-n+1~~~(ii)

We obtained a linear problem of non-negative integer solutions in (ii). The number of solutions to this type of problem are known: \dfrac{[(n)+(m-n+1)]!}{(n)!(m-n+1)!}=\dfrac{(m+1)!}{n!(m-n+1)!}

[I can write the proof if you want]

Now, we just have to calculate the number of forms to permute the men that are dispposed in the row: m!

Multiplying all results:

n!\times\dfrac{(m+1)!}{n!(m-n+1)!}\times m!\\\\&#10;\boxed{\boxed{\dfrac{m!(m+1)!}{(m-n+1)!}}}

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3 years ago
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