Hello AAPL.
This is a really interesting question.
We have a circle is inscribed in a square with an area of 121 square units, then the side of the square is equal to $$Side=\sqrt{121}=11.$$ Since the circle is inscribed, this implies that the circle touches each side of the square (actually in the middle point of each side). Hence, the diameter of the circle is equal to the side of the square, that is to say, $$diameter=11\ \ \ \Rightarrow\ \ \ length\ of\ the\ circle\ =2r\cdot\pi=d\cdot\pi=11\pi.$$ Since the diameter of each marble is 1.5Ï€, then around the circumference can fit $$\frac{11\pi}{1.5\pi}\approx 7.3\ marbles\ \ \ \Rightarrow\ \ \ \ 7\ marbles\ can\ fit.$$ Now, the question asks for how many different combinations of 9 distinct marbles can fit around the circumference of the circle, then we have to compute the number of ways to pick 7 marbles from a group of 9, that is to say, $$9\ C\ 7=C\left(9,7\right)=\frac{9!}{7!\cdot2!}=\frac{9\cdot8\cdot7!}{7!\cdot2}=9\cdot4=36.$$ We have 36 different ways to pick the 7 marbles.
Now, since the marbles are different among them, we have to order them, this can be done with a permutation. Hence, each set of 7 selected balls has 7! different ways to be arranged. Therefore, the number of different combinations of 9 distinct marbles that can fit around the circumference of the circle is given by $$7!\cdot36=181,440.$$ This is why the correct answer is the option D.
I hope it can help you.
