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?

CalculatorUser Apr 12, 2019

#1**0 **

Question was misintinturped

HiylinLink 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

edited by HiylinLink Apr 12, 2019

edited by HiylinLink Apr 12, 2019

edited by HiylinLink Apr 12, 2019

#2

#3**+1 **

Wow I must have really misinturpeted it then because I have no Idea what your talking about so sorry about that incovience

HiylinLink
Apr 12, 2019

#4**+2 **

I'm going to assume that the pairs aren't labeled, and that order within the pairs and triplet don't matter.

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

.Rom Apr 12, 2019