Odd numbers in odd positions

Text

The numbers $1, 2, 3, 4, 5, 6$ are arranged in random order. What is the probability that all the odd numbers occupy odd positions?

Solution

There are $3$ odd numbers ($1,3,5$) and $3$ odd positions ($1^\text{st}, 3^\text{rd}, 5^\text{th}$). We use classical probability.

• Possible cases: all permutations of the $6$ numbers, i.e. $6! = 720$.
• Favorable cases: the $3$ odd numbers in the $3$ odd positions ($3!$ ways) and the $3$ even in the $3$ even positions ($3!$ ways): $3!\cdot 3! = 36$.

Hence $P = \frac{36}{720} = \frac{1}{20} = \boxed{0.05}.$