Ten guests are to be seated in a row of which 3 are ladies. The ladies insist on sitting together while two of the gentlemen refuse to take consecutive seats. In how many ways can the guests be seated?

To make the required arrangements consider 3 ladies as on unit (Name L1, L2, L3). Suppose two gentle who refuse to take consecutive seats be X and Y. Now arrange remaining 5 gentlemen and ladies unit in a row. It can be done in 6! Ways. In each arrangement 3 ladies can be arranged among themselves in 3! Ways. Now arrange X and Y in the gaps between the remaining people. There are 7 gaps. Arrange X and Y in any 2 of 7 gaps. This can be done in $7_{P_2}$ ways.  

The Number of required arrangements  $=6!\times 3!\times 7_{P_2}=720\times 6\times 42=181440$