Is this answer to this one $\frac{3!\cdot12!\cdot3}{15!}=\frac{3}{455}$?

Instead of thinking of groups, think of the number of ways to arrange the 15 distinct people: $$\text{Petruk, Gareng, Bagong, other people}$$such that $A, B,$ and $C$ are in the first three spots. Note that the problem tells us to ignore orders. Thus, ignoring $A, B,$ and $C$, we can now restate the question as the number of ways to arrange the distinct letters $$a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9, a_{10}, a_{11}, a_{12}$$such that the order within the specified groups do not matter. This is given by $\frac{12!}{3!^{2}2!^{3}}$, since there are $12!$ arrangements regarding order, and in each case, we must account for overcounting within individual groups. Similarly, there are $\frac{15!}{3!^{3}2!^{3}}$ total arrangements (now including Petruk, Gareng, and Bagong). Therefore, our probability is $$\frac{\frac{12!}{3!^{2}2!^{3}}}{\frac{15!}{3!^{3}2!^{3}}}={\frac{1}{455}}$$However, we must also consider the fact that Petruk, Gareng, and Bagong can be in any of the three groups, so we multiply the answer by $3$. $\boxed{\frac{3}{455}}$

samrocksnature wrote: Instead of thinking of groups, think of the number of ways to arrange the 15 distinct people: $$\text{Petruk, Gareng, Bagong, other people}$$such that $A, B,$ and $C$ are in the first three spots. Note that the problem tells us to ignore orders. Thus, ignoring $A, B,$ and $C$, we can now restate the question as the number of ways to arrange the distinct letters $$a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9, a_{10}, a_{11}, a_{12}$$such that the order within the specified groups do not matter. This is given by $\frac{12!}{3!^{2}2!^{3}}$, since there are $12!$ arrangements regarding order, and in each case, we must account for overcounting within individual groups. Similarly, there are $\frac{15!}{3!^{3}2!^{3}}$ total arrangements (now including Petruk, Gareng, and Bagong). Therefore, our final probability is $$\frac{\frac{12!}{3!^{2}2!^{3}}}{\frac{15!}{3!^{3}2!^{3}}}=\boxed{\frac{2}{91}}$$ I think your calculation is wrong and you have to consider Petruk,Gareng and Bagong can be in 3 groups not 1 group.Thus,the answer is $3/455$ And it is too easy to be an olympiad problem.

Yay answer confirmed! Assume that a row of $15$ people is partitioned into groups of $3, 3, 3, 2, 2,2$ in that order. Then Petruk, Gareng, and Bagong can be in either of the $3$ groups and can be arranged in $3!$ ways in each group. Also, the other people can be arranged in $12!$ ways. Since there are $15!$ ways to arrange the rows in total, the answer is $\frac{3!\cdot12!\cdot3}{15!}=\frac{3}{455}$.

franzliszt wrote: Yay answer confirmed! Assume that a row of $15$ people is partitioned into groups of $3, 3, 3, 2, 2,2$ in that order. Then Petruk, Gareng, and Bagong can be in either of the $3$ groups and can be arranged in $3!$ ways in each group. Also, the other people can be arranged in $12!$ ways. Since there are $15!$ ways to arrange the rows in total, the answer is $\frac{3!\cdot12!\cdot3}{15!}=\frac{3}{455}$. My solution same with @samrocksnature`s solution i just correct this $$\frac{\frac{12!}{3!^{2}2!^{3}}}{\frac{15!}{3!^{3}2!^{3}}}=\boxed{\frac{1}{455}}$$and i just did multiplication with 3 because Petruk,Gareng and Bagong can be in 3 groups not 1.@samricksnature`s solution doesn`t involve this.

Thx oops Corrected