First slide

Combinations

Question

Number of ways in which 200 people can be divided in 100 couples is

Moderate

A

$\frac{(200)!}{2^{100} (100)!}$
B

$1 \times 3 \times 5 \times \dots \times 199$
C

$(\frac{101}{2}) (\frac{102}{2}) \dots (\frac{200}{2})$
D

$\frac{(200)!}{(100)!}$

Solution

$\frac{(200)!}{\underset{100 times}{\underset{⏟}{2! 2! \dots 2!}} (100)!} = \frac{(200)!}{100! 2^{100}} = 1 \times 3 \times 5 \dots 199 Also, \frac{(200)!}{100! 2^{100}} = (\frac{101}{2}) (\frac{102}{2}) \dots (\frac{200}{2})$

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