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# Combinatorics

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In how many different ways can 3 men and 4 women be placed into two groups of two people and one group of three people if there must be at least one man and one woman in each group?

Apr 12, 2019

#1
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Question was misintinturped

Apr 12, 2019
edited by HiylinLink  Apr 12, 2019
edited by HiylinLink  Apr 12, 2019
edited by HiylinLink  Apr 12, 2019
edited by HiylinLink  Apr 12, 2019
#2
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resolved

Apr 12, 2019
edited by CalculatorUser  Apr 18, 2019
#3
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Wow I must have really misinturpeted it then because I have no Idea what your talking about so sorry about that incovience

$$\text{Let's form the triplet first}\\ \text{There are }\dbinom{3}{1}=3 \text{ ways to choose the male }\\ \text{and }\dbinom{4}{2}=6 \text{ ways to choose the females for a total of 18 ways}\\ \text{next we form a pair. There are }\dbinom{2}{1}=2 \text{ ways to choose the male}\\ \text{and }\dbinom{2}{1}=2 \text{ ways to choose the female for a total of 4 ways}\\ \text{but the two pairs are unlabeled so we have to divide the total arrangments by }2!=2\\ \text{Thus there are }\dfrac{18\cdot 4}{2}=36 \text{ ways to arrange the people as required.}$$