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
For this type of problem, order doesn't matter in which you select the number of people out of the certain group. We use combination to solve the problem.Some notes to know what is going on with this problem:• You want to form a committee of 2 teachers and 5 students to be formed from 7 teachers and 25 students • Then, you select 2 teachers out of 7 without repetition and without considering about the orders of the teachers.• Similarly, you select 5 students out out 25 without repetition and without considering about the orders of the students.Therefore, the solution is (25 choose 5)(7 choose 2) ways, which is equivalent to 1115730 ways to form such committee!
3 students can be chosen from a class of 30 in (30 x 29 x 28) = 24,360 ways.But each group of the same 3 students will be chosen in six different ways.The number of different groups of 3 is 24,360/6 = 4,060 .
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
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
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 was asked
A group of students were