Aclassroom has two rows of 8 seats each. there are 14 students, 5 of whom always sit in the front row and 4 of whom always sit in the back row. in how many ways can the students be seated.

  28,449,792,000

Step-by-step explanation:

There are 8P5 = 6720 ways the five students who sit on the front row can arrange themselves.

There are 8P4 = 1680 ways the four students who sit on the back row can arrange themselves.

In the remaining 7 seats, there are 7P5 = 2520 ways the remaining 5 students can arrange themselves.

The total number of possible arrangements of students (with the given limitations) is ...

  6720·1680·2520 = 28,449,792,000

nPk = n!/(n-k)!

so 8P5 = 8!/3! = 8·7·6·5·4 = 6720

three-fourths of an hour

36÷48=0.75

0.75=

answer: 2/4

step-by-step explanation: