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.
To approach these questions, you can use the fundamental counting principle, which states that if there are "m" ways to do one thing and "n" ways to do another thing, then there are m * n ways to do both things together. Here is how you can apply this principle to each of the questions:
1. In this question, there are 7 people to arrange. First, consider the case where Andrea and Brian are standing together. Treat them as a single unit. Now you have 6 units to arrange: AB, Carol, David, Emmalee, Floyd, Gloria. There are 6! (6 factorial) ways to arrange these 6 units.
Next, consider the case where Andrea and Brian are not standing together. In this case, you can assume that they are separated by the other 5 people and treat each person as a unit. Now you have 3 units to arrange: AB (as a single unit), Carol, David, Emmalee, Floyd, Gloria. There are 3! ways to arrange these 3 units.
Finally, use the fundamental counting principle to combine the two cases: There are 6! + 3! ways to arrange the 7 people if Andrea and Brian must not stand beside one another.
2. In this question, there are 5 people to arrange in a 5-passenger car. However, only 2 of them are willing to drive. So, choose 2 people out of the 5 to be the drivers. This can be done in C(5, 2) ways, which is calculated using the combination formula C(n, r) = n! / (r! * (n-r)!).
Once you have chosen the drivers, you have 5 - 2 = 3 remaining people to arrange in the car. This can be done in 3! ways.
Finally, use the fundamental counting principle to combine the choices for drivers and arrangements of the remaining people: There are C(5, 2) * 3! = 10 * 6 = 60 seating arrangements that can be formed.
3. In this question, there are 6 sandwich fillings, 4 fruit items, and 5 dessert items. To find the total number of lunches, you need to multiply the number of choices for each category.
For the sandwich fillings, there are 6 choices. For the fruit item, there are 4 choices. For the dessert item, there are 5 choices. Use the fundamental counting principle to combine the choices: There are 6 * 4 * 5 = 120 possible lunches that Andrew can have.
Remember to always consider the number of choices for each category and then multiply them together.
I hope this helps! Let me know if you have any further questions.
#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