Peter has 12 relatives (5 man & 7 woman) and his wife also has 12 relatives (5 woman &7 man). They do not have common relatives. They decided to invite 12 guests ,six each of their relatives, such that there are six man and six woman among the guests. How many ways can they choose 12 guests?


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Question Number 120209 by benjo_mathlover last updated on 30/Oct/20

$P e t e r h a s 12 r e l a t i v e s (5 m a n & 7 w o m a n)$ $a n d h i s w i f e a l s o h a s 12 r e l a t i v e s$ $(5 w o m a n & 7 m a n) . T h e y d o n o t$ $h a v e c o m m o n r e l a t i v e s . T h e y d e c i d e d$ $t o i n v i t e 12 g u e s t s, s i x e a c h o f$ $t h e i r r e l a t i v e s, s u c h t h a t t h e r e$ $a r e s i x m a n a n d s i x w o m a n$ $a m o n g t h e g u e s t s . H o w m a n y$ $w a y s c a n t h e y c h o o s e 12 g u e s t s ?$
	Answered by bemath last updated on 30/Oct/20

	$\underset{k = 0}{\sum^{5}} {[(\begin{matrix} 5 \\ k \end{matrix}) (\begin{matrix} 7 \\ 6 - k \end{matrix})]}^{2} = 267 148$
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