I've been thinking about this one problem for 2 days. The book does not give enough examples for me to figure it out. HEre is the problem.
A panel is being organized. You need to arrange a list of participants 4 administrators and 4 students. A) in how many ways can you list them if administrators must sit together in a group and students in a group? B) In how many ways can you list them if you must alternate students and administrators?
The answer on the back of the book says its 2*4!*4! = 1152. However it does not specified if this applies to part A) part B) or both. Plus I have no clue how they got this arrangement. Please provide how you came up with your answer. And yes its part of my homework but I just have no clue on how to solve this one.
2007-09-10 02:59:50 · 2 answers · asked by mr_gees100_peas 6 in Science & Mathematics ➔ Mathematics
A) P=2*(4! * 4!)=2*24*24=2*576=1152 ; -in 1152 ways
only students can sit in 24 diferent ways, admins. in 24 different ways, by multylpying you get the way they could sit together in one planel, but only considering one group for example at the left side, the other at the right side, by mulitplying with two you get both positions for admins. and student
b) V=4!*4!=576, depending on who is the first , an admin os student we multiply the result with 2 , and we get 1152 ways
2007-09-10 03:17:55 · answer #1 · answered by Anonymous · 0⤊ 0⤋
4! is the number of permutations of students
4! is the number of permutations of administrators
2 is the option of starting with a student or an administrator
part A and B are the same. Ultimately part A is worded in a confusing manner.
2007-09-10 10:09:01 · answer #2 · answered by Sugar Shane 3 · 0⤊ 1⤋