Placing Objects: Englishman, French, Turks In how many ways can you place 6 English, 7 French and 10 Turkish men in a line so that each Englishman is between a French me and a Turkish, and no French men is next to a Turkish man.
 A: A partial answer only. 
Start with the six Englishmen. Then we need to fill the gaps between each two consecutive Englishmen, as well as the spaces at both ends of the line. That's seven spaces in all. 
In each space, there must be only Frenchmen or only Turks. Furthermore, according to the conditions, they must alternate. Thus the possible patterns are FETEFETEFETEF and TEFETEFETEFET, where each E represents an Englishman, each F one or more Frenchmen, and each T one or more Turks.
For each of these two patterns, you must count the number of ways to distribute the ten Turks and the seven Frenchmen among the spaces allotted to them.
A: For simplicity, assume that we only care about the nationality of the individuals, and not their individual identity.
First line up the 10 Turks in a row, leaving 2 outside gaps and 9 inside gaps.
Every group of Frenchmen in an inside gap requires exactly 2 Englishmen as a buffer, and every group of Frenchmen in an outside gap requires exactly 1 Englishman as a buffer. 
Since there are 6 Englishmen, this gives us two cases to consider:
1) If the Frenchmen are in 3 inside gaps, then they cannot be in either outside gap (since all 6 Englishmen are also on the inside), so there are $\binom{9}{3}$ ways to choose the inside gaps and $\binom{6}{2}$ ways to distribute them in the 3 gaps.
2) If the Frenchmen are in 2 inside gaps, then they must also occupy both outside gaps (since each outside gap contains an Englishman); so there are $\binom{9}{2}$ ways to choose the 2 inside gaps and $\binom{6}{3}$ ways to distribute them in the 4 gaps.
Therefore there are a total of $\binom{9}{3}\binom{6}{2}+\binom{9}{2}\binom{6}{3}$ ways to place them in a line under these conditions
(if we consider all people of each nationality the same).
