One must first consider the number of ordered ways that 4 students may be chosen, and then divide that by the number of ways the four may have been ordered to create a number of distinct groups.
For the orders, there are 30 choices for the first, 29 for the second, 28 for the third, and 27 for the fourth. However, there are a 4 ways within the four to pick the first, 3 ways for the second, 2 for the third, and 1 left for the fourth. The answer may be expressed by the following:
30 x 29 x 28 x 27
______________
1 x 2 x 3 x 4
which evaluates to 27,405 committees.
40 x 39 x 38 x 37 = 2193360
The first member chosen can be any one of 1,514 students.The second member chosen can be any one of the remaining 1,513 students.The third member chosen can be any one of the remaining 1,512 students.So there are (1,514 x 1,513 x 1,512) ways to choose three students.But for every group of three, there are (3 x 2 x 1) = 6 different orders in which the same 3 can be chosen.So the number of `distinct, unique committees of 3 students is(1514 x 1513 x 1512) / 6 = 577,251,864
The first member chosen can be any one of 4,463 students.The second member chosen can be any one of the remaining 4,462 students.The third member chosen can be any one of the remaining 4,461 students.So there are (4,463 x 4,462 x 4,461) ways to choose three students.But for every group of three, there are (3 x 2 x 1) = 6 different orders in which the same 3 can be chosen.So the number of `distinct, unique committees of 3 students is(4463 x 4462 x 4461) / 6 = 14,805,989,111
18
To determine how many 3-member committees can be formed from a group of 18 students, you can use the combination formula: (C(n, r) = \frac{n!}{r!(n-r)!}), where (n) is the total number of students and (r) is the number of members in the committee. In this case, (n = 18) and (r = 3). Thus, the calculation is (C(18, 3) = \frac{18!}{3!(18-3)!} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816). Therefore, you can form 816 different 3-member committees from the group of 18 students.
15
40 x 39 x 38 x 37 = 2193360
The first member chosen can be any one of 1,514 students.The second member chosen can be any one of the remaining 1,513 students.The third member chosen can be any one of the remaining 1,512 students.So there are (1,514 x 1,513 x 1,512) ways to choose three students.But for every group of three, there are (3 x 2 x 1) = 6 different orders in which the same 3 can be chosen.So the number of `distinct, unique committees of 3 students is(1514 x 1513 x 1512) / 6 = 577,251,864
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.
7 women in a group of 11 people 7/11
The first member chosen can be any one of 4,463 students.The second member chosen can be any one of the remaining 4,462 students.The third member chosen can be any one of the remaining 4,461 students.So there are (4,463 x 4,462 x 4,461) ways to choose three students.But for every group of three, there are (3 x 2 x 1) = 6 different orders in which the same 3 can be chosen.So the number of `distinct, unique committees of 3 students is(4463 x 4462 x 4461) / 6 = 14,805,989,111
7 women in a group of 11 people 7/11
The house rules committee. APEX
The committee of Ten was a group of educators in 1892 that played a big role in the standardization of American high schools. There were different philosophies that the group of ten aimed to resolve.
A study group committee is often referred to as an academic committee. Its primary purpose is to facilitate peer-learning, group study, and academic support among its members.
A group of students were
yes