Answer : Since the committee of 5 is to be formed from 6 gents and 4 ladies. (i) Forming a committee with at least 2 ladies Here the possibilities are
The number of ways they can be selected = 186 ways= 4C2 X6C3 + 4C3X6C2 + 4C4X6C1 (ii) The number of ways in this case is
The total ways are = 4C0 X6C5 + 4C1X6C4 + 4C2 X 6C3 = 186 ways. Answer: (2) 246 Permutation and combination are the ways to represent a group of objects by selecting them in a set and forming subsets. It defines the various ways to arrange a certain group of data. When we select the data or objects from a certain group, it is said to be permutations, whereas the order in which they are represented is called combination. PermutationIn mathematics, permutation relates to the act of arranging all the members of a set into some sequence or order. A permutation is the choice of r things from a set of n things without replacement and where the order matters. nPr = (n!) / (n-r)! CombinationThe combination is a way of selecting items from a collection, such that the order of selection does not matter.  Solution:The possible arrangements of 6 men and 4 women to form a committee of 5 members can be done in different ways, such as: 1 Lady +4 gents 2 Ladies +3 gents 3 Ladies +2 gents 4 Ladies +1 Gents 5 Ladies +0 gents Therefore, total number of arrangements = (4C1×6C4)+(4C2×6C3)+(4C3×6C2)+(4C4×6C1) = 60+120+60+6 = 246 Permutation and Combination
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The number of groups with exactly $k$ of the four women and $5 - k$ of the six men is $$\binom{4}{k}\binom{6}{5 - k}$$ so the number of groups with exactly $k$ women is \begin{align*} \sum_{k = 1}{4} \binom{4}{k}\binom{6}{5 - k} & = \binom{4}{1}\binom{6}{4} + \binom{4}{2}\binom{6}{3} + \binom{4}{3}\binom{6}{2} + \binom{4}{4}\binom{6}{1}\\ & = 4 \cdot 15 + 6 \cdot 20 + 4 \cdot 15 + 1 \cdot 6\\ & = 60 + 120 + 60 + 6\\ & = 246 \end{align*} as you found more easily by subtracting the number of groups of five people with no women from the number of groups of five people that can be formed without restriction. Suppose the women are Angela, Brenda, Charlotte, and Denise and that the men are Edward, Frank, George, Henry, Ivan, and Jeffrey. By designating a particular woman as the representative of the women in the group, you count each group the number of times as the number of women in the group. For instance, consider the group consisting of the four women and Edward. You count it in each of the following ways: \begin{array}{c c} \text{designated woman} & \text{additional members of the group}\\ \hline \text{Angela} & \text{Brenda, Charlotte, Denise, Edward}\\ \text{Brenda} & \text{Angela, Charlotte, Denise, Edward}\\ \text{Charlotte} & \text{Angela, Brenda, Denise, Edward}\\ \text{Denise} & \text{Angela, Brenda, Charlotte, Edward} \end{array} Notice that $$\color{red}{\binom{1}{1}}\binom{4}{1}\binom{6}{4} + \color{red}{\binom{2}{1}}\binom{4}{2}\binom{6}{3} + \color{red}{\binom{3}{1}}\binom{4}{3}\binom{6}{2} + \color{red}{\binom{4}{1}}\binom{4}{4}\binom{6}{1} = 504$$ where the factor $\color{red}{\binom{k}{1}}$ is the number of ways of designating one of the $k$ women in the group as the representative of the women in the group. |
How many ways can a committee of 5 members consisting of 4 men and 1 woman be selected from 6 men and 5 women?
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