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

y-y1=4(x-x1)

step-by-step explanation:

this is point-slope form since you only have 4 as the slope it goes in place of m

hope this : )

the answer is 48m^2!

step-by-step explanation: