There are 10 different sets of teachers which can be combined with 4 different sets of students, so 40 possible committees.
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.
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 .
8/33
KWL charts are a great tool for students and teachers. It can be used with any subject to help students with comprehension. In the first column students write what they know about the subject. The second column (W) students write what they want to know about a subject. The third column (L) is where students write what they learned. KWL charts are a great way to help students learn not just what they already know but help them to see what they are learning.
in my opinion, 5% of the world are teachers bcoz 5 is my lucky number and it helps me all the time. Abid Shah
There are 10560 possible committees.
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.
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!
To determine the number of different committees that can be formed from 11 teachers and 48 students, we need to clarify the size of the committee and whether there are any restrictions on the selection. If we assume that any combination of teachers and students can be chosen without restrictions, the total number of possible combinations is (2^{11} \cdot 2^{48} = 2^{59}). This accounts for every possible subset of teachers and students, including the empty committee.
To form a committee of 2 teachers from 5, we use the combination formula ( \binom{n}{r} ), where ( n ) is the total number and ( r ) is the number chosen. The number of ways to choose 2 teachers from 5 is ( \binom{5}{2} = 10 ). For the 4 students from 10, the number of ways is ( \binom{10}{4} = 210 ). Therefore, the total number of different committees is ( 10 \times 210 = 2100 ).
no teachers don't tickle students
The nouns in the sentence are students and teachers.
why should teachers challenge 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.
1 : 3
I am on the housing committee. I am on the committee of teachers who regulate the rules for school. The city council has a committee meeting once a month.
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 .