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.
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In how many ways can a committee of 5 members be formed from [#permalink]
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Question Stats: Hide Show timer StatisticsIn how many ways can a committee of 5 members be formed from 4 women and 6 men such that at least 1 woman is a member of the committee? (A) 112 (B) 156 (C) 208 (D) 246 (E) 252 _________________
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Re: In how many ways can a committee of 5 members be formed from [#permalink]
Carcass wrote: In how many ways can a committee of 5 members be formed from 4 women and 6 men such that at least 1 woman is a member of the committee? (A) 112 (B) 156 (C) 208 (D) 246 (E) 252 Because the committee needs to consist of 5 people:Choose one girl and 4 dudes,choose 2 girls and 3 dudes,etc....4C1 * 6C4 +4C2 * 6C3 +4C3 * 6C2 +4C4 * 6C1 = 246So D.
Intern Joined: 27 Jun 2019 Posts: 40
Re: In how many ways can a committee of 5 members be formed from [#permalink]
Carcass wrote: In how many ways can a committee of 5 members be formed from 4 women and 6 men such that at least 1 woman is a member of the committee? (A) 112 (B) 156 (C) 208 (D) 246 (E) 252 There are 10C5 = 252 ways to pick a 5 member committee out of 10 people (this includes committees with at least one woman and committees with no women at all). To get the number of committees with at least one woman, we should subtract the number of committees with no women from the total number of possible committees. There are 6C5 = 6 ways to pick the 5 member committee from men only. Thus, the number of committees where at least 1 woman is a member = 252 - 6 = 246. Answer D.
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3. In how many ways can a committee of members be formed from male and female students? 4. From a standard deck of cards, how many ways can cards be drawn? |
In how many ways can a committee of 5 members be formed from 10 male and 8 female students?
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