Im on the chapter of permutations/combinations/fundamental counting principle
need help with 2 questions
1. emma is arranging her books in piles on her desk. she has 5 math books, 3 english books, and 4 science books. how many ways can she arrange them if she wants to keep the subject areas 2gether?
I did 12!/4!3!5!, but that's wrong, it should be 360 :|
2. gordon has some spare change on his desk, he got a nickel, penny and quarter. assumin that he uses at least 1 coin, how many diff sums r possible?
^ i did 3P1 * 3P1 * 3P1, but the answer is supposed to be 12 :|
There are 3! ways to arrange the subjects
For each such arrangement, there are 5! * 3! * 4! ways to arrange the books.
That would be 3!3!4!5! = 103680
No idea how 360 is the answer.
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C(3,1)+C(3,2)+C(3,3) = 3+3+1
No idea how 12 is the answer. The coins can be picked in the following ways:
p,n,q, pn,pq,nq, pnq
The order does not matter.
1. To find the number of ways Emma can arrange her books while keeping the subject areas together, you need to use the concept of permutations.
Breaking down the problem:
- Emma has 5 math books. There are 5! (5 factorial) ways to arrange them among themselves.
- Emma has 3 English books. There are 3! ways to arrange them among themselves.
- Emma has 4 science books. There are 4! ways to arrange them among themselves.
Since Emma wants to keep the subject areas together, we need to find the number of ways Emma can arrange the three groups (math books, English books, and science books) among themselves. This can be done in 3! ways.
The correct way to calculate the total number of arrangements is by multiplying these values together: 5! * 3! * 4! * 3! = 360.
So, your initial calculation of 12!/4!3!5! was incorrect because you didn't consider the permutations within each subject area.
2. To find the number of different sums possible using a nickel, penny, and quarter, you should consider the combinations of coins that can be taken together. In this case, you can use the concept of combinations.
Breaking down the problem:
- Gordon has 3 different coins: nickel, penny, and quarter.
- Assuming he uses at least 1 coin, you need to find the number of combinations of these coins.
There are three possibilities:
- Using only one coin: This can be done in 3 ways.
- Using two coins: This can be done by selecting 2 coins out of 3, which is represented as 3C2. Calculating this combination:
- 3C2 = 3! / (2! * (3-2)!) = 3.
- Using all three coins: This can be done in 1 way.
To find the total number of different sums possible, add up the possibilities for each case: 3 + 3 + 1 = 7.
Therefore, your calculation of 3P1 * 3P1 * 3P1 was incorrect. Instead, you should calculate the combinations as explained above, resulting in an answer of 7, not 12.