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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?
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.
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.
