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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?
Wiki User
∙ 6y ago
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.
Wiki User
∙ 6y ago
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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 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 do teachers use fractions?
Teachers teach fractions to their students.
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 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
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 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 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 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) ]
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!
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.
Will teachers layoffs effect on students grades?
absoloutly! teachers have different teaching skills, and if too many teachers are replaced, then then the teachers may teach differently, which is confusing to the kids, and they may not do well on tests. STUDENTS NEED CONSISTANCY!
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why should teachers challenge students
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Every teacher and school is different. Some may or may not recommend students use the site. However, many teachers use this site.
