10 guests are to be seated in a row. 3 of them are to be seated together. of the remaining 7, 2 do not wish to sit side by side. find the number of possible arrangements.

Question

anonymous · Answer

Do you know if its combination or permutation?

anonymous · Answer

First we should figure out how many seating arrangements there are with the 3 people sitting together, regardless of where the other two people are. If we think of the three people as 1 object (so they cant separate) then  we realize this is the same question as "how many ways can you permute 8 objects?"  Its 8, because the three people are now 1 object, and there are 7 others to worry about.  That makes 8! ways.

Now, instead of counting the ways of having the two others apart, we should put them together, and subtract from 8!. So now the other two people are also 1 object.  With the three people as 1, the other two also as 1, and the remaining 5 people, we are asking how many ways can we permute 7 objects.  Thats 7!. 

Hence the number of ways we can have the 3 people sitting together, with the other two apart is 8! - 7!.

anonymous · Answer

just a query: 
for the first part i guess the number of combination in which 3 people always sit together the possible combination will be 8!x3! as the 3 people can also sit together in 3 factorial ways.
similarly for the second part it would be 7!x3!x2!

kropot72 · Answer

@digitalmonk My result is 3!(8! - 2!7!), which appears to be the same as yours.

anonymous · Answer

oops! forgot about the number of ways the 3 people can move around and 2 people can move around >.<