Question 1:
There are 12 students in a class. Find the number of ways that 12 studenets can take 3 different tests if 4 students are to take each test.
Question 2
Find the number of ways 12 students can be partitioned into 3 teams A, B, C, so that each team contain 4 students.
I don't get what is the difference between the two. For the first one, the book uses C(12,4) * C(8,4) = 34650. I use the permutation method, which is bascially the same 12!/(4!4!4!) = 34650
For the second question, the books tells me to think that A is in team A and you have to chose three different people to join team A, resulting into C(12,3). The same logic applies to team B: C(7,3) and that left 4 for team C.
Therefore the answer was C(12,3) * C(7,3) = 5925.
THe book tells me if you use the partition method, all you do is 34650/6 = 5925 because 3! is an ordered partition. (what is an ordered partition anyway?)
Ah, the wonderful world of combinatorics! Let's see if we can make some sense out of it.
In the first question, we have 12 students and we want to divide them into three groups of 4 for each test. The approach you used with permutations is correct. You have 12 choices for the first student in the first group, then 11 choices for the second student, and so on until you have 9 choices left for the fourth student in the first group. This gives us a total of 12 * 11 * 10 * 9 = 11,880 ways to choose the first group.
But hold on, we don't actually care about the order in which we choose the students! That's where combinations come in. We have 4 students in the first group, and any arrangement of those 4 students would still form the same group. So, we divide by the number of possible arrangements of those 4 students, which is 4! (4 factorial, or 4 * 3 * 2 * 1). Similarly, we divide by 4! for the second and third groups, resulting in your answer of 34,650 (or C(12,4) * C(8,4)).
Now, let's move on to the second question. This time we want to partition the 12 students into three teams, A, B, and C, with 4 students in each team. Here, we are not concerned with the order in which the teams are chosen. To count the number of ways, we can start with the total number of ways to choose the students for each team.
For Team A, we choose 4 students from the original 12, so there are C(12,4) ways to do that. After choosing the first 4 students, we have 8 students left, and we choose another 4 for Team B, giving us C(8,4) ways. Finally, we have 4 students left, who will automatically form Team C. So, the total number of ways to partition the students is C(12,4) * C(8,4).
But wait! We have counted each partition multiple times because the order of choosing the teams matters. If we consider the three teams as Team A, Team B, and Team C, we would count the partition (A,B,C) differently from (A,C,B), even though they are essentially the same. To correct for this, we divide the total number of partitions by 3! = 6 (which represents the number of different ways to arrange the teams). This finally gives us the answer of 5,925 (or C(12,4) * C(8,4) / 6).
So, in summary, the difference between the two questions lies in the order: the first question is concerned with the order in which students are chosen for each group, while the second question is concerned with the order of the teams themselves. That's why we need to consider permutations in the first question and divide by a factor of 6 in the second question.
And as for what an ordered partition is, it simply means that the order of the groups (teams A, B, and C in this case) matters. In an ordered partition, each group or subset is assigned a specific position or order, whereas in an unordered partition, only the members of the groups or subsets matter, not their order.
For Question 1, the number of ways that 12 students can take 3 different tests if 4 students are to take each test can be calculated using the combination formula. The formula is C(n, r) = n! / (r!(n-r)!), where n represents the total number of students and r represents the number of students per test.
By applying this formula, we can calculate the number of ways as follows:
C(12,4) * C(8,4) = (12! / (4!(12-4)!)) * (8! / (4!(8-4)!))
= (12! / (4!8!)) * (8! / (4!4!))
= (12*11*10*9*8*7*6*5*4! / (4!8!)) * (8*7*6*5*4! / (4!4!))
= (12*11*10*9) / (4*3*2*1) * (8*7*6*5) / (4*3*2*1)
= 34650
So, there are 34650 ways for 12 students to take 3 different tests, with 4 students per test.
For Question 2, we need to find the number of ways 12 students can be partitioned into 3 teams, A, B, and C, with 4 students in each team. This can be calculated by using the concept of ordered partitions.
An ordered partition means that the order in which the teams are formed matters. In this case, we first choose 4 students for team A, then choose 4 students for team B, and then the remaining 4 students automatically form team C.
To calculate the number of ways to form the teams using the combination formula, we can apply the concept of ordered partitions.
C(12,3) * C(7,3) = (12! / (3!(12-3)!)) * (7! / (3!(7-3)!))
= (12! / (3!9!)) * (7! / (3!4!))
= (12*11*10) / (3*2*1) * (7*6*5*4) / (3*2*1)
= 220 * 35
= 7700
So, there are 7700 ways for 12 students to be partitioned into 3 teams, A, B, and C, with 4 students in each team.
Therefore, the book's answer of 34650/6 = 5925 is correct because dividing by 6 accounts for the ordered partition.
Question 1:
To find the number of ways that 12 students can take 3 different tests, with 4 students in each test, you can use the combination formula.
In this case, we want to select 4 students out of 12 for the first test, and then select another 4 students out of the remaining 8 for the second test. So, you can calculate it as follows:
C(12,4) * C(8,4) = 34650
The combination formula C(n, r) calculates the number of ways to choose r items from a group of n items, without regard to the order of the items.
Alternatively, you used the permutation method to calculate the same result. The permutation formula P(n, r) calculates the number of ways to arrange r items from a group of n items, considering the order of the items. In this case, the formula would be:
12! / (4! * 4! * 4!) = 34650
The exclamation mark denotes factorial, which means multiplying a number by all the positive integers less than it down to 1.
Question 2:
To find the number of ways to partition 12 students into 3 teams, with 4 students in each team, you can use the combination formula again.
The book suggests approaching this problem by thinking about selecting students for each team separately.
For team A, you need to select 4 students out of the 12. This can be calculated as C(12, 4).
Similarly, for team B, you need to select 4 students out of the remaining 7 (12 students minus the 4 from team A). This can be calculated as C(7, 4).
Finally, the remaining 4 students will automatically be in team C.
So, the number of ways to partition the students into teams A, B, and C would be:
C(12, 4) * C(7, 4) = 5925
Now, for understanding the concept of an ordered partition:
An ordered partition means that the teams have a specific order or sequence. In this case, team A is selected first and then team B. When using the ordered partition method, you need to account for the order of the teams.
In this problem, since you are just interested in the number of ways to partition the students into the teams, which doesn't consider the order of the teams, you can divide the result obtained from the combination method by the number of all possible ordered partitions, which is 3!.
3! (read as "three factorial") is equal to 3 * 2 * 1 = 6.
So, by dividing the result obtained from the combination method by 6, you get the same result:
34650 / 6 = 5925
Thus, using the ordered partition method, you can simplify the calculation.