0
How many different committees can be formed from 12 teachers and 36 students if the committee consists of 4 teachers and 3 students?
Wiki User
∙ 11y ago
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.
Wiki User
∙ 11y ago
Add your answer:
Earn +20 pts
How many different committees can be formed from 9 teachers and 49 students if the committee consists of 2 teachers and 4 students?
Possibilities are (9 x 8)/2 times (49 x 48 x 47 x 46)/24 = 366,121,728/48 =7,627,536 different committees.
How many different committees can be formed from 9 teachers and 41 students if the committee consist of 4 teachers and 2 student?
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.
Is this a permutation In how many different ways could a committee of 5 students be chosen from a class of 25 students?
6,375,600
How many different ways can 1 committee of 5 students be selected from a class of 25?
53,130 ways.
How many ways are there to select six students from a class of twenty five to hold six different executive positions on a committee?
That depends. Are all the positions equal, will they all just be members of the committee or will there be a chairman, vice-chairman, secretary, treasurer, etc.? Assuming all are equal positions, then order of selection does not matter, then (25x24x23x22x21x20)/(6x5x4x3x2x1) = 177,100
How many different committees can be formed from 6 teachers and 33 students if the committee consists of 3 teachers and 2 students?
There are 10560 possible committees.
How many different committees can be formed from 9 teachers and 49 students if the committee consists of 2 teachers and 4 students?
Possibilities are (9 x 8)/2 times (49 x 48 x 47 x 46)/24 = 366,121,728/48 =7,627,536 different committees.
How many different committees can be formed from 10 teachers and 30 students if the committee consists of 2 teachers and 2 students?
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 .
How many ways can a committeee of 6 be chosen from 5 teachers and 4 students if the committee must includes three teachers and three students?
There are 10 different sets of teachers which can be combined with 4 different sets of students, so 40 possible committees.
How many different committees can be formed from 9 teachers and 41 students if the committee consist of 4 teachers and 2 student?
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.
Is this a permutation In how many different ways could a committee of 5 students be chosen from a class of 25 students?
6,375,600
How many different ways can 1 committee of 5 students be selected from a class of 25?
53,130 ways.
How many ways can a committee of 6 be chosen from 5 teachers and 4 students if all are equally eligible?
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) ]
A committee of 4 students will be selected from a list that contains 6 grade 9 students and 8 grade 10 What is the expected number of grade 10s on the committee?
8
Who Founded the students nonviolent coordinating committee?
Ellen Baker
Why to join placement committee in college?
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.
How many committees consisting of 3 female and 5 male students can be selected from a group of 5 female and 8 male students?
The number of ways is (5C3)*(8C5) = [(5*4*3)/(3*2*1)]*[(8*7*6)/(5*4*3)] = 10*56 = 560 committees.
