There are 10560 possible committees.
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 .
For the first spot, you can choose any one of 5 students. For the second spot, you can choose any one of the remaining 4 students. For the third spot, you can choose any one of the remaining 3 students. etc. So the answer is: 5x4x3x2x1 = 120
19*18*17*16*15 = 1,395,360
I Dont Know [210]
To find the number of different groups (a) within a larger group (b), use the formula b!/a!.(b-a)! where the ! sign indicates "factorial". In your problem b = 21 and a = 5 so we have 21!/(5!.16!) this simplifies to 21.20.19.18.17/5.4.3.2 cancelling leaves 21.19.3.17 ie 20349
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 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.
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.
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) ]
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.
The number of ways is (5C3)*(8C5) = [(5*4*3)/(3*2*1)]*[(8*7*6)/(5*4*3)] = 10*56 = 560 committees.