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 .
There are 10 different sets of teachers which can be combined with 4 different sets of students, so 40 possible committees.
To calculate the number of ways a committee of 6 can be chosen from 5 teachers and 4 students, we use the combination formula. The total number of ways is given by 9 choose 6 (9C6), which is calculated as 9! / (6! * 3!) = 84. Therefore, there are 84 ways to form a committee of 6 from 5 teachers and 4 students if all are equally eligible.
Wit?
TARC AFA students?
There are 11880 ways.
There are 10560 possible committees.
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.
6,375,600
53,130 ways.
I think the answer might surprise you! Any 4 from 12 is (12 x 11 x 10 x 9)/(4 x 3 x 2) ie 495; Any 3 from 36 is (36 x 35 x 34)/(3 x 2) ie 7140 These must be multiplied as each teacher set can be combined with each student set giving a total of (deep breath) 3,534,300.
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.
To calculate the number of ways a committee of 6 can be chosen from 5 teachers and 4 students, we use the combination formula. The total number of ways is given by 9 choose 6 (9C6), which is calculated as 9! / (6! * 3!) = 84. Therefore, there are 84 ways to form a committee of 6 from 5 teachers and 4 students if all are equally eligible.
8
Ellen Baker
You can join the placement committee in college to help influence were students are placed. If you are a student this will help your voice be heard.