1.
[tex]\displaystyle\\12\cdot\binom{7}{2}\cdot\binom{5}{4}=12\cdot\dfrac{7!}{2!5!}\cdot\dfrac{5!}{4!}=12\cdot\dfrac{6\cdot7}{2}\cdot5=1260[/tex]
2.
[tex]\displaystyle\\\binom{9}{6}+\binom{9}{7}+\binom{9}{8}+1=\dfrac{9!}{6!3!}+\dfrac{9!}{7!2!}+\dfrac{9!}{8!}+1=\dfrac{7\cdot8\cdot9}{2\cdot3}+\dfrac{8\cdot9}{2}+9+1=\\=7\cdot4\cdot3+4\cdot9+10=130[/tex]
