UNO High School Problem of the Week Competition
Problem 3
There are 10 open windows in a house. In how many different orders can all of the windows
be closed and locked, provided that each window must be closed prior to becoming locked?
Solution 1
Let's assume that we can do any action on any window in a random order. Each of the 20! permutaions
is equally probably. For any fixed window, the probability that it will be closed before it is locked
is 1⁄2, so the probility that all windows will be closed and locked properly is 1⁄210. Therefore the number
of correct permutations must be 20!⁄210, which is 2,375,880,867,360,000.
Solution 2
Each closing sequence for the windows can be described as a sequence containing two 1s, two 2s, two
3s, ..., two 10s where the first occurrence means that the window was closed and the second
occurrence means that the window was locked. The number of these sequences can be described by
20C2*18C2* ... * 2C2. By the factorial definition of the nCr function, this becomes
20!⁄(2!*(20-2)!)*18!⁄(2!*(18-2)!)* ... *2!⁄(2!*(2-2)!). We note that the denominator of one term contains the
numerator of the next term, so the equation simplifies to 20!⁄(2!)10.