Permutations and Combinations Question (1 Viewer)

Ambility · Nov 17, 2015

Nine people are to be seated at a thin rectangular table with 4 people on one side and five on the other. How many different ways can these nine people be seated if:

Mary and John must sit on the same side

Mary and John must sit on opposing sides

Drsoccerball · Nov 17, 2015

Ambility said:
Nine people are to be seated at a thin rectangular table with 4 people on one side and five on the other. How many different ways can these nine people be seated if:

Mary and John must sit on the same side

Mary and John must sit on opposing sides

$(i) $ Fix Mary and John on one side, lets start with the 4 seats side=A and the 5 seats side=B. $$

$A=\binom{7}{2}4!\binom{5}{5}5!$

$B= \binom{7}{3}5!\binom{4}{4}4!$

$\therefore A+B = same side$

$(ii) \binom{9}{5}5!4!- $ same side $$

$\binom{9}{5}5!4!-A-B$

calamebe · Nov 29, 2015

I had different working but got the same answer so I'll post my working as well. I didn't really explain it well so if anyone wants me to explain it I can. Here's the working:
http://m.imgur.com/BFyRvzg

Permutations and Combinations Question (1 Viewer)

Ambility

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Drsoccerball

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calamebe

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