Answer:
252 different combinations
Step-by-step explanation:
There are 12 men who may be elected to the council. If there are only 5 members of the council (the order doesn't matter), then there are
[tex]C^{10}_5\\ \\=\dfrac{10!}{5!(10-5)!}\\ \\=\dfrac{10!}{5!\cdot 5!}\\ \\=\dfrac{5!\cdot 6\cdot 7\cdot 8\cdot 9\cdot 10}{5!\cdot 5!}\\ \\=\dfrac{6\cdot 7\cdot 8\cdot 9\cdot10}{1\cdot 2\cdot 3\cdot 4\cdot 5}\\ \\=7\cdot 4\cdot 9\\ \\=252[/tex]
different combinations of council members.
