#1 First consider all the ways where A&B are together. This is equivalent to permuting 6 students (AB,C,D,E,F,G): 6! = 720.
But, we have to double this, since AB and BA are two possible ways that A&B can be together.
There are 7! ways to permute the 7 students.
7! - 2*6! = 720 - 240 = 480
#2.
2 ways to pick the driver
4!=24 ways to seat the others.
Looks like 48 ways to seat them.
What was your logic?
#3.
2 breads
6 fillings
so, 2*6*5 = 60 different sandwiches
(once the 1st sandwich is made, there are only 5 choices for the other)
4 fruits
5 desserts
so, 60*4*5 = 1200 different lunches
I'm having a hard time understanding the fundamental counting principal and I don't even know where to start on some of the questions.Here are a few questions I'm having trouble with.
1.Andrea, Brian, Carol, David, Emmalee, Floyd, and Gloria are to stand in a line for a photograph. If they each face the photographer, in how many ways can they be arranged if Andrea and Brian must not stand beside one another?(This one I have no idea how to approach)
2.How many seating arrangements can be formed by 5 people traveling in a 5-passenger car if only 2 of the 5 people are willing to drive? 48-16=32
3.Andrew's mother always makes his lunch for him. She packs one brown-bread sandwich, one white-bread sandwich, one fruit item, and one dessert item. Andrew likes ham, cheese, salami, chicken, turkey, and jam as sandwich fillings. He likes apples, grapes, bananas, and pears, and he likes Oreo cookies, chocolate cake, fudge, butterscotch pudding, and O'Henry clusters for dessert. How many lunches can Andrew have if his mother puts a different filling in each sandwich?(This one I don't know how to do either)
4.
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