PLS HELP A classroom of students can be divided
into two, three, and five groups, with no
students left over. How many students are
likely in the class? Explain why.

In how many ways can 12 students be divided into 3 groups of 4 for a group project, if the order of the groups and the arrangement of the students in a group are immaterial?

The question you ask fits into the category of the so-called multinomial. In you have n objects (in this case 12 spots) and if there are k1 of type 1, k2 of type 2, and so on up to km of type m with ∑mj=1kj=n (in this case there are three types of spots each with four students), then the total number of possible arrangements is given by:

n!k1!×k2!×…×km!

In your problem, the result would be:

12!4!×4!×4!

However, the multinomial is appropriate when the m types (in your problem, the groups) are distinguishable. For example, if you were picking groups to work on 3 different problems, then having a set of four people assigned to group 1 would be different than assigning the same set of four people assigned to work on group 2. Since there are 3 groups, there are 3!=6 otherwise identical ways the groups can be arranged. Since we counted all of these identical rearrangements even though we were not supposed to, we have overcounted by a factor of 6 . That gives a final answer of:

12!3!×(4!)3

32/80  

91/140

the last one 54 doesn't go into 45. 54/100 ?

step-by-step explanation:

answer: 193

explanation:

200-7 = 193