Edwin is spending the afternoon house hunting. There are 6 properties on his list, and he has time to see 2 of them. How many sequences of 2 houses are possible?
Question
Edwin is spending the afternoon house hunting. There are 6 properties on his list, and he has time to see 2 of them. How many sequences of 2 houses are possible?
Solution
To solve this problem, we can use the concept of permutations in combinatorics. A permutation is an arrangement of objects in a specific order. The number of permutations of n objects taken r at a time is given by the formula:
P(n, r) = n! / (n-r)!
where "!" denotes factorial, which is the product of all positive integers up to that number.
In this case, Edwin has 6 properties (n=6) on his list and he has time to see 2 of them (r=2). So we can substitute these values into the formula:
P(6, 2) = 6! / (6-2)!
= 720 / 24
= 30
So, there are 30 different sequences of 2 houses that Edwin could potentially visit.
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