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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 different committees can be formed from 12 teachers and 36 students if the committee consists of 4 teachers and 3 students?
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.
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 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 different committees can be formed from 6 teachers and 49 students if the committee consists of 4 teachers and 4 students?
To determine the number of different committees that can be formed with 4 teachers from 6 and 4 students from 49, we use combinations. The number of ways to choose 4 teachers from 6 is given by ( \binom{6}{4} ), and the number of ways to choose 4 students from 49 is ( \binom{49}{4} ). Thus, the total number of different committees is ( \binom{6}{4} \times \binom{49}{4} ). Calculating this gives ( 15 \times 194580 = 2918700 ) different committees.
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 different committees can be formed from 12 teachers and 36 students if the committee consists of 4 teachers and 3 students?
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.
How many 3 member committed can you choose from a group of 18 students?
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.
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
