\section*{Problem 6 - 310936}

For the sequence in which the nine letters \(A, B, C, D, E, F, G, H, J\) can be arranged from left to right, the following seven conditions are required:

\(A\) should be to the left of \(B\), \(A\) should be to the left of \(C\), \(A\) should be to the left of \(D\), \(E\) should be to the left of \(F\), \(E\) should be to the left of \(G\), \(E\) should be to the left of \(H\), \(E\) should be to the left of \(J\).

How many different sequences, in which these seven conditions are met, are there in total?

Hints:

In each of the mentioned sequences, each of the nine letters should appear exactly once.

The phrase " \(X\) to the left of \(Y\) " does not exclude the possibility that other letters may stand between \(X\) and \(Y\).
}

We divide the nine letters into two groups. Group 1 contains \(A, B, C\) and \(D\), Group 2 the remaining five letters. The conditions compare only the positions of letters within the same group, so any distribution of the nine positions to the two groups yields a correct

arrangement of all nine letters, provided that the conditions are all met within both groups.

There are a total of 6 possible arrangements in Group 1 that satisfy the three conditions: \(A\) must be on the far left (among these four elements), while the order of the other three letters in this group can be arbitrary, thus allowing all \(3!=6\) variants. Similarly, for Group 2, there are \(4!=24\) possible arrangements (since \(E\) must be on the far left among these, while the remaining four letters of Group 2 can be arranged arbitrarily).

The nine positions can be distributed to the two groups in \(\binom{9}{4}=\frac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 1}=126\) ways (without considering the order of the distribution of the positions). For each such distribution, there are 6 possible arrangements of the elements of Group 1 among themselves on the positions of this group and 24 possible arrangements of the elements of Group 2 on their positions, resulting in a total of

\[
126 \cdot 6 \cdot 24=126 \cdot 144=18144
\]

possible arrangements.