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There are 10560 possible committees.
There are 46,150 possible committees.
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.
Such a committee is typically called a "select committee" or "special committee." These committees are formed to address specific issues, conduct investigations, or carry out functions that are not within the purview of standing committees. They may be temporary and disband after their purpose is fulfilled, often leading to recommendations for future legislation.
Members of Congress usually choose committee chairperson on the basis of seniority. Nominations are made by the majority party caucus or conference.
Yes, a subcommittee can be a committee of the whole. While it is possible, it is usually stupid. The purpose of a subcommittee is to study an issue in depth, to bring back a report, and present the research to back it up. If the subcommittee does its job, it keeps the committee's discussion from being a pooling of ignorance. Usually committees divide up work among subcommittees. If you are going to have a subcommittee be a committee of the whole, why have a subcommittee?
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 .
To determine how many 5-person committees can be formed from nine people, you can use the combination formula ( C(n, k) = \frac{n!}{k!(n-k)!} ), where ( n ) is the total number of people, and ( k ) is the number of people to choose. In this case, ( n = 9 ) and ( k = 5 ). Thus, the number of ways to select the committee is ( C(9, 5) = \frac{9!}{5!(9-5)!} = \frac{9!}{5!4!} = 126 ). Therefore, there are 126 different 5-person committees possible.
This is a combination problem; use the nCr formula. nCr = 15! / ((15-4)!*4!) nCr = 15! /(11!*4!) nCr = 1365
There are 10 different sets of teachers which can be combined with 4 different sets of students, so 40 possible committees.
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In an experiment containing a number of possible variables only one of these should be altered in each individual experiment. Otherwise it is not possible to determine which variable is responsible for a particular change.