Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).
Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2
49=16+x^2
x^2= 33
x= square root of 33
Let's write an inequality, such as follows: x < sqrt(50) < y. Square both sides of the equation. We get x^2 < 50 < y^2. Obviously, x is between 7 and 8. Also notice, that for integers a,b, (ab)^2/b^2, equals a^2. So let's try values, like 7.1. Using the previous fact, (7.1)^2, equals (71)^2/100. So, (7.1)^2, equals 50.41. Thus, our number is between 7 and 7.1. We find, with a bit of experimentation, that the square root of 50, is 7.07.