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!
32
84 students and six teachers.Students per teacher = (number of students) / (number of teachers)= 84 / 6= 14
18
20 x 19 x 18/3 x 2 = 1,140 groups
A school has 18 classes with 35 students in each class. In order to reduce class size to 30, how many new classes must be formed?
There are 10560 possible committees.
There are 84 different combinations possible for the committee of 6, taken from4 students and 5 teachers.1.- The committee with 4 students has 4C4 number of combinations of 4 students out of 4 and 5C2 number of combinations of 2 teachers out of 5 to be combined with. The product of these two give the number of different combinations possible in a committee formed by 4 students and 2 teachers.2.- The committee with 3 students has 4C3 number of combinations of 3 students out of 4 and 5C3 number of combinations of 3 teachers out of 5 to be combined with. The product of these two give the number of different combinations possible in a committee formed by 3 students and 6 teachers.3.- The committee with 2 students has 4C2 number of combinations of 2 studentsout of 4 and 5C4 number of combinations of 4 teachers out of 5 to be combined with. The product of these two give the number of different combinations possible in a committee formed by 2 students and 4 teachers.4.- The committee with 1 student has 4C1 number of combinations of 1 student out of 4 students and 5C5 number of combinations of 5 teachers out of 5 to be combined with. The product of these two give the number of different combinations possible in a committee formed by 1 student and 5 teachers.We now add up all possible combinations:4C4∙5C2 + 4C3∙5C3 + 4C2∙5C4 + 4C1∙5C5 = 1(10) + 4(10) +6(5) + 4(1) = 84There are 84 different combinations possible for the committee of 6, taken from4 students and 5 teachers.[ nCr = n!/(r!(n-r)!) ][ n! = n(n-1)(n-2)∙∙∙(3)(2)(1) ]
There are (10 x 9)/2 = 45 different possible pairs of 2 teachers. For each of these . . .There are (30 x 29)/2 = 435 different possible pairs of students.The total number of different committees that can be formed is (45 x 435) = 19,575 .
Possibilities are (9 x 8)/2 times (49 x 48 x 47 x 46)/24 = 366,121,728/48 =7,627,536 different committees.
There are 10 different sets of teachers which can be combined with 4 different sets of students, so 40 possible committees.
You can select 4 of the 9 teachers in any order, and for each of those selections you can select 2 of the 41 students in any order. This is two combinations → number_of_ways = ₉C₄ + ₄₁C₂ = 9!/((9-4)!4!) + 41!((41-2)!2!) = 126 + 820 = 946 different committees.
56 divided by 3 is 18 remainder 2.
There are 25 teachers and 487 students.
the answer is 56 divided by 3 is 18 remainder 2
32
86 students, 34 teachers pop: 120
84 students and six teachers.Students per teacher = (number of students) / (number of teachers)= 84 / 6= 14