How many combinations are possible if 1 man and 1 woman refuse to work with each other?

Mathematics

1 answer:

trapecia [35]3 years ago

6 0

If it is a combination of 2 then there is only 1 but if it is a combination of 1 then there is 2, hope this help!!

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Sidana [21]

<h2>Hello!</h2>

The answer is:

$\frac{14}{12}=\frac{7}{12}*2\\\\\\\frac{21}{12}=\frac{7}{12}*3\\\\\\\\\frac{28}{12}=\frac{7}{12}*4\\\\\\\\\frac{35}{12}=\frac{7}{12}*5$

A multiple of a number is the repeated sum of itself, for example:

Multiple of the number 1 are: 1, 2, 3, 4, 5 and so...

Multiple of the number 3 are: 3, 6, 9, 12, 15 and so...

From the example we have:

Multiple of the number 3 are: Itself, (3+3), (3+3+3), (3+3+3+3), (3+3+3+3+3) and so...

So,

Multiple of $\frac{7}{12}$ are:

$\frac{14}{12}=\frac{7}{12}*2=\frac{7}{12}+\frac{7}{12}\\\\\\\frac{21}{12}=\frac{7}{12}*3=\frac{7}{12}+\frac{7}{12}+\frac{7}{12}\\\\\\\\\frac{28}{12}=\frac{7}{12}*4=\frac{7}{12}+\frac{7}{12}+\frac{7}{12}+\frac{7}{12}\\\\\\\\\frac{35}{12}=\frac{7}{12}*5=\frac{7}{12}+\frac{7}{12}+\frac{7}{12}+\frac{7}{12}+\frac{7}{12}$