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I understand the concept of combinations and permutations. However, I am not getting how to apply the formulas. I believe understanding exactly how to do this would help. A club has 24 members. a. In how many ways can 4 officers, a president, vice-president, secretary and treasurer be chosen from the members of the club? b. In how many ways can a 4-person committee be chosen from the members of the club?
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Answer Hint: This question is from permutation and combinations. In this question, we are going to use the formula \[{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}\]. This is the formula for when we have to select r things from n things.First, we will see how many things we have to select from the total. After that, we will solve the question using the above formula. Complete step by step answer: Let us solve this question.This is a question of permutation and combinations.Let us first know what permutations and combinations.A number of permutations (order matters) of n things taken r at a time: \[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\]A number of combinations (order does not matter) of n things taken r at a time:\[{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}\]So, in this question, we have to find that in how many ways, we can select a committee of five members from a group of 10 people.So, in this question, the process of choosing the persons does not matter which means the order of choice does not matter. So, we are going to use the formula for this question is \[{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}\]We are choosing 5 persons from 10 persons where order does not matter.Then, the numbers of ways are \[\dfrac{10!}{(10-5)!5!}\]As we know that \[n!=n\times (n-1)\times (n-2)\times (n-3)\times ........\times 2\times 1\]Then, \[\dfrac{10!}{(10-5)!5!}\] can be written as \[\dfrac{10!}{(10-5)!5!}=\dfrac{10!}{5!\times 5!}=\dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{\left( 5\times 4\times 3\times 2\times 1 \right)\times \left( 5\times 4\times 3\times 2\times 1 \right)}\]Which is also can be written as\[\dfrac{10!}{(10-5)!5!}=\dfrac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 1}=\dfrac{10}{5}\times 9\times \dfrac{8}{4}\times 7\times \dfrac{6}{3\times 2}=2\times 9\times 2\times 7\times 1=252\]Hence, \[\dfrac{10!}{(10-5)!5!}=252\]Therefore, the number of ways of selecting a committee of 5 members from a group of 10 persons is 252. Note: For solving this type of question, we should have a better knowledge of permutation and combinations. And, also remember the formulas of permutation and combination. Make sure that no calculation mistakes have been done in solving the question. Otherwise, the solution will be wrong. For example, \[\dfrac{10!}{(10-5)!5!}=\dfrac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 1}=\dfrac{10}{5}\times 9\times \dfrac{8}{4}\times 7\times \dfrac{6}{3\times 2}=2\times 9\times 2\times 7\times 1=252\]In the above, mistakes can be made during calculations. |
From a 20 member club how many ways can a committee of five students be selected
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