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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.
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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 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 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 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.
A school committee consists of 2 teachers and 4 students. The number of different committees that can be formed from 5 teachers and 10 students is?
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 ).
Which came first committees of correspondence formed or second continental congress formed?
the committees of correspondence formed are came first
What committees are temporary house and senate committees formed to study key issues?
Conference Committees
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 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.
Which committees were formed In 1772 to improve communication between colonies?
Committees of correspondence
Which committees were formed in 1772 to improve communication between the colonies?
Committees of correspondence
