in how many ways can a class of nineteen students be divided into three sets so that six students are in the first set, seven students are in the second, and six are in the third?

The 12 students can be lined up in 12! ways.

Put in dividers after every fourth student.

For example:

Given ABCDEFGHIJKL, we would put in dividers to get ABCD|EFGH|IJKL.

We have overcounted, though, in two different ways. Let's examine the resulting problems.

One problem is with regard to the three separate groupings appearing multiple times. In the example above, we would be counting EFGH|ABCD|IJKL a second time. Since there are three groups, we need to divide by 3! to avoid this sort of overcounting.

The second problem is that, within each of the three groups, we have four people who are going to appear multiple times. In the example above, we would be counting CDAB|EFGH|IJKL a second time. For each of the groupings, we need to divide by 4! to avoid this sort of overcounting. This means dividing by 4!4!4! to amelioriate the second issue.

Then our answer becomes:

12!3!4!4!4!=5,775

Step-by-step explanation:

hope it help brainliest pls

12cm

the answer is d

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204