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A company hires 13 new people who has to be distributed over 2 cities --- 5 in New York and 8 in Tokyo. Bob and Jim are in this group of new employees. How many ways can these 13 people be assigned so that Bob and Jim do not work in the same city ?

Please help!

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closed as unclear what you're asking by Shailesh, iadvd, hardmath, user91500, TastyRomeo Feb 1 '17 at 9:50

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HINTS

  1. Each valid way assigns other 11 employees into groups of $5-1=4$ and $8-1=7$.
  2. For each of these assignments, we can have Jim with 4 and Bob with 7 or vice versa.
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Let's re-frame this problem - a method which tends to be useful in many types of math problems (particularly in my opinion statistics problems).

Consider the group without Bob and Jim, we will deal with them later!

You are then left with 11 employees, of which 4 need to go to New York, and 7 need to go to Tokyo. This is a much simpler problem! Simply take $11 \choose 4$ to pick the 4 which need to go to New York! (Note it would have been equally acceptable to take $11 \choose 7$ which is equal to $11 \choose 4$!)

If you are unable to see why this is the case, then I would recommend reading up on some more basic combinatorial problems (i.e. books on a bookshelf!)

Now, you can either have Bob join New York, or Bob join Tokyo (this forces Jim to Tokyo or New York, respectively). This is the same as asking "How many ways can you have two people, work in two different cities?" The answer is of course $2$.

So, we know that, there are $11 \choose 4$ ways to arrange the other $11$ individuals (which equals $330$) and then there are 2 ways to arrange Bob and Jim. $2 \times 330 = 660$.

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