QUESTÃO 05: Calcule. a) (} 8 6 ) b) (} 7 3 ) c) (} 11 5 )

Vamos corrigir e detalhar os cálculos para garantir a precisão:<br /><br />a) \(\binom{8}{6}\)<br /><br />\[<br />\binom{8}{6} = \frac{8!}{6!(8-6)!} = \frac{8!}{6!2!} = \frac{8 \times 7 \times 6!}{6! \times 2 \times 1} = \frac{8 \times 7}{2 \times 1} = 28<br />\]<br /><br />b) \(\binom{7}{3} + \binom{7}{4}\)<br /><br />\[<br />\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5 \times 4!}{3! \times 4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35<br />\]<br /><br />\[<br />\binom{7}{4}frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5 \times 4!}{4! \times 3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} =<br />\]<br /><br />\[<br />\binom{7}{3} + \binom{7}{4} = 35 + 35 = 70<br />\]<br /><br />c) \(\binom{11}{5} + \binom{12}{5} + \binom{11}{6}\)<br /><br />\[<br />\binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11!}{5!6!} = \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6!}{5! \times 6!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462<br />\]<br /><br />\[<br />\binom{12}{5} = \frac{12!}{5!(12-5)!} = \frac{12!}{5!7!} = \frac{12 \times 11 \times 10 \times 9 \8 \times 7!}{5! \times 7!} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792<br />\]<br /><br />\[<br />\binom{11}{6} = \frac{11!}{6!(11-6)!} = \frac{11!}{6!5!} = \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6!}{6! \times 5!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462<br />\]<br /><br />\[<br />\binom{11}{5} + \binom{12}{5} + \binom{11}{6} = 462 + 792 + 462 = 1716<br />\]<br /><br />Portanto, as respostas corrigidas são:<br /><br />a) 28<br /><br />b) 70<br /><br />c) 1716