First slide

Combinations

Question

The value of $E = \frac{(1 + 17) (1 + \frac{17}{2}) (1 + \frac{17}{3}) â¦ (1 + \frac{17}{19})}{(1 + 19) (1 + \frac{19}{2}) (1 + \frac{19}{3}) â¦ (1 + \frac{19}{17})}$ is

Moderate

A

1
B

$^{36} C_{17}$
C

$\frac{2}{19}$
D

$^{36} C_{18}$

Solution

We have
$(1 + k) (1 + \frac{k}{2}) (1 + \frac{k}{3}) â¦ (1 + \frac{k}{n})$
$\begin{array}{l} = \frac{(1 + k) (2 + k) (3 + k) â¦ (n + k)}{(2) (3) â¦ (n)} = \frac{(n + k)!}{k! n!} \\ =^{n + k} C_{k} \end{array}$
Thus, both the numerator and the denominator of $E$ equals
$36_{C_{17}} = 36_{C_{19}} â$
$â´ E = 1 .$

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