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Question Number 120209 by benjo_mathlover last updated on 30/Oct/20
$${Peter}\:{has}\:\mathrm{12}\:{relatives}\:\left(\mathrm{5}\:{man}\:\&\:\mathrm{7}\:{woman}\right) \\ $$ $${and}\:{his}\:{wife}\:{also}\:{has}\:\mathrm{12}\:{relatives} \\ $$ $$\left(\mathrm{5}\:{woman}\:\&\mathrm{7}\:{man}\right).\:{They}\:{do}\:{not} \\ $$ $${have}\:{common}\:{relatives}.\:{They}\:{decided} \\ $$ $${to}\:{invite}\:\mathrm{12}\:{guests}\:,{six}\:{each}\:{of} \\ $$ $${their}\:{relatives},\:{such}\:{that}\:{there} \\ $$ $${are}\:{six}\:{man}\:{and}\:{six}\:{woman}\: \\ $$ $${among}\:{the}\:{guests}.\:{How}\:{many} \\ $$ $${ways}\:{can}\:{they}\:{choose}\:\mathrm{12}\:{guests}? \\ $$
Answered by bemath last updated on 30/Oct/20
![Σ_(k=0) ^5 [ ((5),(k) ) ((( 7)),((6−k)) )]^2 = 267 148](Q120210.png)
$$\:\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\:\:\left[\begin{pmatrix}{\mathrm{5}}\\{{k}}\end{pmatrix}\begin{pmatrix}{\:\:\:\:\mathrm{7}}\\{\mathrm{6}−{k}}\end{pmatrix}\right]^{\mathrm{2}} =\:\mathrm{267}\:\mathrm{148} \\ $$