# Probability – What Is the Chance All Choices Are Different

permutationprobabilityself-study

Suppose everyone (7 people) chooses independently and randomly out of 7 choices.

What are the odds that no two people have picked the same option?

I feel like the answer is $$5040/823543,$$ but I am rusty with anything math related.

There are $$7$$ ways for one person to make a choice. Assuming they choose uniformly, which means no person favors any choice over any other, each option therefore has a chance of $$1/7.$$ (This follows directly from the probability axioms, which assert the sum of all seven equal chances equals $$1.$$)

When $$n$$ people independently make choices, the very definition of independence means that any given array of choices has a chance of $$1/7\times \cdots \times 1/7 = 1/7^n.$$

An array consisting of all distinct choices denotes a permutation of those choices. There are $$7$$ permutations of one choice, $$7\times (7-1)$$ permutations of two choices (because the second cannot agree with the first), $$7\times (7-1)\times (7-2)$$ permutations of three choices (two possibilities are excluded from the first choice), and so on. The number of permutations of all $$7$$ options therefore is $$7\times 6\times \cdots \times 1 = 7! = 5040.$$

Since all permutations of the choices are distinct, the probability axioms tell us to add the chances of all these permutations. This amounts to multiplying the common chance of $$1/7^n$$ by the number permutations, giving

$$\frac{1}{7^7} \times 7! = \frac{5040}{823543} \approx 0.006119899$$

as stated in the question.