A1. In the bowl, we have 10 marbles of three different colors: blue, yellow, and green. We can arrange them in a row in 360 different ways. What is the maximum number of blue marbles that can be in the bowl?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
A1. Let the number of blue, yellow, and green balls be denoted by $m, r$, and $z$ respectively, where $m + r + z = 10$. These balls can be arranged in a row in $\frac{10!}{m!r!z!}$ different ways. Therefore, $\frac{10!}{m!r!z!} = 360$. We can rearrange this equation to $10 \cdot 9 \cdot \ldots \cdot (m+1) = 360 \cdot r! \cdot z!$. From this, it follows that $10 \cdot 9 \cdot \ldots \cdot (m+1) \geq 360 = 10 \cdot 9 \cdot 4$, so $m+1 \leq 8$ or $m \leq 7$. For example, if $m=7, r=2$, and $z=1$, then $\frac{10!}{m!r!z!} = \frac{10!}{7!2!1!} = \frac{10 \cdot 9 \cdot 8}{2} = 360$. Therefore, the bag can contain at most 7 blue balls. The correct answer is $D$.