if order does not matter then, (23x22x21x20x19)/(5x4x3x2x1) = 33,649
20 x 19 x 18/3 x 2 = 1,140 groups
The number of different groups of students that can be formed from 16 students depends on the size of the groups being formed. If you are looking for all possible combinations of groups of any size (from 1 to 16), you can use the formula for combinations. The total number of combinations would be (2^{16} - 1) (subtracting 1 to exclude the empty group), which equals 65,535 different groups. If you specify a particular group size, the calculation would be different.
To determine how many ways 4 students can be chosen from a class of 12 and assigned different tasks, we first select 4 students from the 12, which can be done in ( \binom{12}{4} ) ways. Then, we can assign the 4 different tasks to these students in ( 4! ) (24) ways. Therefore, the total number of ways to choose the students and assign the tasks is ( \binom{12}{4} \times 4! = 495 \times 24 = 11,880 ).
To determine how many different teams of 9 can be chosen from 12 students, we use the combination formula (C(n, k) = \frac{n!}{k!(n-k)!}), where (n) is the total number of students and (k) is the number of students to choose. Here, (n = 12) and (k = 9). Thus, the calculation is (C(12, 9) = C(12, 3) = \frac{12!}{3! \cdot 9!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220). Therefore, there are 220 different teams of 9 that can be chosen from 12 students.
two
6,375,600
30C8 = 5,852,925
20 x 19 x 18/3 x 2 = 1,140 groups
There are 11880 ways.
The number of different groups of students that can be formed from 16 students depends on the size of the groups being formed. If you are looking for all possible combinations of groups of any size (from 1 to 16), you can use the formula for combinations. The total number of combinations would be (2^{16} - 1) (subtracting 1 to exclude the empty group), which equals 65,535 different groups. If you specify a particular group size, the calculation would be different.
To determine how many ways 4 students can be chosen from a class of 12 and assigned different tasks, we first select 4 students from the 12, which can be done in ( \binom{12}{4} ) ways. Then, we can assign the 4 different tasks to these students in ( 4! ) (24) ways. Therefore, the total number of ways to choose the students and assign the tasks is ( \binom{12}{4} \times 4! = 495 \times 24 = 11,880 ).
There are 10 different sets of teachers which can be combined with 4 different sets of students, so 40 possible committees.
Well, honey, there are 30 students in the class, and you want to choose a group of 3. So, you're looking at a classic combination situation. The formula for combinations is nCr = n! / r!(n-r)!, so in this case, it's 30C3 = 30! / 3!(30-3)! = 4060 ways to choose those 3 lucky students. It's like picking the winning lottery numbers, but with fewer tears and more math.
There are 247 groups comprising 2 or more students.
To determine how many different teams of 9 can be chosen from 12 students, we use the combination formula (C(n, k) = \frac{n!}{k!(n-k)!}), where (n) is the total number of students and (k) is the number of students to choose. Here, (n = 12) and (k = 9). Thus, the calculation is (C(12, 9) = C(12, 3) = \frac{12!}{3! \cdot 9!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220). Therefore, there are 220 different teams of 9 that can be chosen from 12 students.
Any 5 from 7 is (7 x 6)/2 ie 21.
2.33333333