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How many ways can a committee of 2 teachers and 5 students to be formed from 7 teachers and 25 students?
For this type of problem, order doesn't matter in which you select the number of people out of the certain group. We use combination to solve the problem.
Some notes to know what is going on with this problem:
• You want to form a committee of 2 teachers and 5 students to be formed from 7 teachers and 25 students • Then, you select 2 teachers out of 7 without repetition and without considering about the orders of the teachers.
• Similarly, you select 5 students out out 25 without repetition and without considering about the orders of the students.
Therefore, the solution is (25 choose 5)(7 choose 2) ways, which is equivalent to 1115730 ways to form such committee!
What else can I help you with?
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.
There are 4 married couples in a club A 3 member committee must be formed from among them such that no married couple is part of the committee In how many ways can this committee be formed?
32
A class trip consists of 84 students and 6 teachers how many students per teacher are there?
84 students and six teachers.Students per teacher = (number of students) / (number of teachers)= 84 / 6= 14
How many different 3 member teams can be formed from a group of 6 students?
18
How many different groups of 3 students can be formed if there are 20 students in the class?
20 x 19 x 18/3 x 2 = 1,140 groups
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 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.
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 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 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 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 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.
There were 56 people on a bus both students and teachers there are 3 times as many students as teachers how many teachers how many teachers are on the bus?
56 divided by 3 is 18 remainder 2.
How many teachers and students are in Armstrong elementary school in Reston VA?
There are 25 teachers and 487 students.
How many teachers on the bus if 56 people on a bus-both students and teachers. There are 3 times as many students as teachers?
the answer is 56 divided by 3 is 18 remainder 2
There are 4 married couples in a club A 3 member committee must be formed from among them such that no married couple is part of the committee In how many ways can this committee be formed?
32
How many students are there at stanbridge academy?
86 students, 34 teachers pop: 120
