Guide 4 - Page 73 - Combinatorics Strategy

[dghosh2602]

G, M, P, J, B and C are supposed to sit in 6 different seats. But M and J won't sit together, how many different arrangements are possible?

So, I know I can calculate the possibilities w/o constraints and subtract the possibilities with constraints and get the answer as 6! - 5! to get 480. What I am trying to think is in terms of the following (which works for the rest of the problems in the chapter, but I can't figure out why the model doesn't fit this situation)

I have 6 possibilities for Seat 1 and I assume its M, so possibilities = 6.
Now, I have 4 possibilities for Seat 2 because J won't sit with M, so possibilities = 4
Now, I have 4 possibilities again for Seat 3 because M is back in the group, so possibilities = 4
And then finally 3 and 2, so possibilities = 3 * 2

Multiplying, 6 * 4 * 4 * 6 = 576 possibilities.

Where am I formulating this wrongly?

Thanks in advance

[tim]

Okay, first, 6!-5! does not equal 480. You have to multiply the 5! by 2 because J and M could go in either order. This gives you 5!*2=240 ways for J and M to sit together. NOW subtract that from 720 and you have the 480 you need..

As for the other approach, if you assume M is seated in seat 1 you DO NOT have 6 possibilities anymore, you have 1 possibility. Then when you get to seat 3, you CANNOT put M there because you already put M in seat 1. So what you have to do is figure out all the ways to place M in seat 1, which is different from the ways to put G in seat 1 for instance, so now you have to add up a bunch of cases and... yeah, you can probably see how that gets really complicated really quickly. :)

If you see a combinatorics problem that looks that difficult, you've missed the trick and you need to step back and see if there is a simpler way..