A group consisting of 5 girls and 7 boys plays "wedding": they select (in the following order) a couple, a registrar, two flower girls each with a gentleman, and two witnesses. How many ways can this "wedding party" be formed if 3 of the girls have 1 brother each present, and the couple cannot be siblings, and the registrar cannot be a sibling of either of the couple? - Does the result change if the registrar is chosen last?
The number of candidates for the position of the chief bridesmaid (hereafter abbreviated as: $A$) varies depending on whether the person selected for the couple $J$ is participating in the game with a sibling or not. If so, it must also be considered whether both tickets are of this type or only one. The number of possibilities for the $K$ flower-bearing couples is influenced by whether $A$ is a girl or a boy. Due to these considerations, the number of possible arrangements for the entire wedding party is determined by breaking it down into several cases.

Let's call those who are boys without a sibling $F$, girls without a sibling $L$, and those who participate with a sibling, boys $B$ and girls $H$. Then $J$ can be of 4 types:

$$
\begin{array}{cccc}
F L & F H & B L & B H
\end{array}
$$

and the number of such possibilities is respectively

$$
(7-3)(5-3)=4 \cdot 2 \quad 4 \cdot 3 \quad 3 \cdot 2 \quad 3 \cdot(3-1)=3 \cdot 2
$$

because there are 4, 2, 3, and 3 individuals in these types, and in the $B H$ type pair, one $B$ cannot be paired with one $H$.

For a rule-compliant $A$, it is sufficient to know in the following whether it is a boy or a girl, briefly: $f$ or $l$ type. Considering this, there are $4 \cdot 2=8$ types for the combination of $J$ and $A$. From $f$ and $l$, the numbers in the above types are respectively

$$
\begin{array}{llll}
6+4 & 5+4 & 6+3 & 5+3
\end{array}
$$

so the number of possibilities in the 8 types is:

$\begin{array}{llll}F L f: 4 \cdot 2 \cdot 6, & F H f: 4 \cdot 3 \cdot 5, & \text { BLf }: 3 \cdot 2 \cdot 6, & \text { BHf }: 3 \cdot 2 \cdot 5 ; \\ F L l: 4 \cdot 2 \cdot 4, & F H l: 4 \cdot 3 \cdot 4, & \text { BLl }: 3 \cdot 2 \cdot 3, & \text { BHl }: 3 \cdot 2 \cdot 3 .\end{array}$

The total number of possibilities for the four types under (2) is $48+60+36+30=174$. In every such case, the remaining participants are chosen from 5 boys and 4 girls. For the first $K$, there are $5 \cdot 4$ possibilities, and for the second, from the remainder, $(5-1)(4-1)=4 \cdot 3$ possibilities, and for the two, half of the product of these: $5 \cdot 4 \cdot 4 \cdot 3 / 2=120$. Indeed, any two specific $K$s can be selected in two orders, but we do not consider these two possibilities different. Thus, for the selection of the first 7 individuals, with a boy $A$, $174 \cdot 120=20880$ possibilities arise.

Similarly, for the types under (3), the continuation of $32+48+18+18=116$ possibilities is the same. From 6 boys and 3 girls, similarly to the above, the two $K$s can be formed in $6 \cdot 3 \cdot 5 \cdot 2 / 2=90$ ways, so with a girl $A$, $116 \cdot 90=10440$ possibilities arise for the selection of the first 7 individuals.

There are no exclusionary criteria for the selection of witnesses, and the $31320$ possibilities found so far all end the same way: from the remaining 5 individuals, 2 can be chosen in $5 \cdot 4 / 2=10$ ways. Thus, the number of possible arrangements for the entire wedding party is $N=313200$.

The order of selection does not affect the number of possibilities, as the order does not appear among the exclusionary criteria. At every stage, we have considered every allowed possibility and excluded every disallowed one. With a different order of selection, we would determine the same number of possibilities through a different grouping.

Endre Miklóssy (Szolnok, Verseghy F. g. IV. o. t.)

Remarks. I. We can reach the sum of the numbers under (1), (2), and (3) more quickly if we do not form types but subtract the number of possibilities that are not allowed to be considered.

From 5 girls and 7 boys, $5 \cdot 7=35$ pairs can be formed, and 3 of these sibling pairs do not meet the criteria for $J$, so the number of properly formed $J$s is $35-3=32$. (This is the sum under (1).)

With the remaining 4 girls, $4 \cdot 32=128$ possibilities are promised for the selection of $J$ and a girl $A$. But this way, before every $H-A$, the brother would also appear as a groom, in 4 ways, with the other 4 girls. Therefore, $3 \cdot 4=12$ cases must be deleted, and $128-12=116$ arrangements meet the rule. (This is the sum under (3).)

Similarly, after the selection of $J$, with 6 boys who can be $A$, $6 \cdot 32=192$ arrangements can be thought of, but $3 \cdot 6=18$ do not meet the criteria, because none of the three $B$s can be the $A$ of the sister and the 6 $J$s formed in sequence with the other 6 boys. The remaining $192-18=174$ possibilities meet the criteria. (This is the sum under (2).)

Mária Katona (Budapest, Szilágyi E. lg. II. o. t.)

2. If we determine in advance that the order of selection of $A$ is not significant, then the consideration of this selection can be inserted at the most favorable place in the calculation for this purpose. Since $A$ is only limited by the selection of $J$, the order $J, A$ is appropriate. If $A$ is selected last, more cases would need to be considered depending on whether the sibling of one or both members of $J$ has been selected in the meantime as a flower-bearing couple or witness, or not.