Respostas

2013-10-01T16:43:46-03:00
Basta usar o fórmula da combinação, onde a combinação de n elementos (todos disponíveis), fazendo de k em k elementos (pegamos k dentre os n disponíveis fazemos todas as combinações). Então a fórmula será:

C_{n,k}  = \frac{n!}{k! . (n - k)!}

a)  \frac{C_{n,k}}{C_{n,n-k}} . C_{n,n} = \frac{\frac{n!}{k! . (n - k)!}}{\frac{n!}{(n - k)! . (n - (n - k))!}} . \frac{n!}{n! . (n - n)!}

\frac{C_{n,k}}{C_{n,n-k}} . C_{n,n} = \frac{n!}{k! . (n - k)!} . \frac{(n - k)! . (n - (n - k))!}{n!} . \frac{n!}{n! . (n - n)!}

\frac{C_{n,k}}{C_{n,n-k}} . C_{n,n} = \frac{1}{k! . 1} . \frac{1 . (n - n + k)!}{1} . \frac{n!}{n! . 1!}

\frac{C_{n,k}}{C_{n,n-k}} . C_{n,n} = \frac{(0 + k)!}{k!} . \frac{n!}{n!}

\frac{C_{n,k}}{C_{n,n-k}} . C_{n,n} = \frac{k!}{k!} . 1

\frac{C_{n,k}}{C_{n,n-k}} . C_{n,n} = 1

b)  \frac{C_{5,k} . (5 - k) . k!(4 - k)!}{30} = \frac{\frac{5!}{k! . (5 - k)!} . (5 - k) . k!(4 - k)!}{30}

\frac{C_{5,k} . (5 - k) . k!(4 - k)!}{30} = \frac{\frac{5 . 4 . 3 . 2 .1}{k! . (5 - k)(4 - k)!} . (5 - k) . k!(4 - k)!}{30}

\frac{C_{5,k} . (5 - k) . k!(4 - k)!}{30} = \frac{\frac{120}{1 . (5 - k)(4 - k)!} . (5 - k) . 1(4 - k)!}{30}

\frac{C_{5,k} . (5 - k) . k!(4 - k)!}{30} = \frac{\frac{120}{1 . 1 . (4 - k)!} . 1 . 1 . (4 - k)!}{30}

\frac{C_{5,k} . (5 - k) . k!(4 - k)!}{30} = \frac{\frac{120}{1 . 1 . 1} . 1 . 1 . 1}{30}

\frac{C_{5,k} . (5 - k) . k!(4 - k)!}{30} = \frac{120}{30}

\frac{C_{5,k} . (5 - k) . k!(4 - k)!}{30} = 4

c)  \frac{C_{n + 1, n} + n + 1}{C_{n + 1,1}} = \frac{\frac{(n + 1)!}{n! . (n + 1 - n)!} + n + 1}{\frac{(n + 1)!}{1! . (n + 1 - 1)!}}

\frac{C_{n + 1, n} + n + 1}{C_{n + 1,1}} = (\frac{(n + 1)!}{n! . (n + 1 - n)!} + n + 1) . (\frac{1! . (n + 1 - 1)!}{(n + 1)!})

\frac{C_{n + 1, n} + n + 1}{C_{n + 1,1}} = (\frac{(n + 1)!}{n! . 1!} + n + 1) . (\frac{1! . n!}{(n + 1)!})

\frac{C_{n + 1, n} + n + 1}{C_{n + 1,1}} = \frac{(n + 1)! + n!(n + 1)}{(n + 1)!}

\frac{C_{n + 1, n} + n + 1}{C_{n + 1,1}} = \frac{(n + 1)! + (n + 1)!}{(n + 1)!}

\frac{C_{n + 1, n} + n + 1}{C_{n + 1,1}} = \frac{2 . (n + 1)!}{(n + 1)!}

\frac{C_{n + 1, n} + n + 1}{C_{n + 1,1}} = \frac{2 . 1}{1}

\frac{C_{n + 1, n} + n + 1}{C_{n + 1,1}} = 2