I have a question about combinatorics.

Here is the question:

A waiting area outside the doctor's office contains a row of 7 chairs. In how many different ways can a man, a woman and a boy occupy 3 of the chairs such that:

a. the man and the boy seated in adjoining chairs

b. all three seated in adjoining chairs

My attempts:

a. Since the man and the boy must be seated in adjoining chairs, so we can conclude that only 2 groups sits in 3 chairs here, so the number of ways is:

P(6,2) * 2= 60 ways

b. Because three people must be seated near to each others, so we can group them such that:

number of ways of sitting: P(5,1)

Because there is 3 people, so the total number of ways of sitting is: 3*P(5,1)= 15 ways

After that, I doubt about my answers. Are my solutions is true or I need to improve that?


  • 1
    $\begingroup$ For the second, there are $3!$ ways to arrange the people once the adjacent chairs have been chosen. So it is $30$, not $15$. $\endgroup$ – André Nicolas Sep 2 '15 at 5:24
  • $\begingroup$ I think you did everything correct except multiplying by 3 in the last step. You are correct that there are 5 spots the trio could choose, but how many ways can you arrange the threesome? $\endgroup$ – turkeyhundt Sep 2 '15 at 5:24
  • $\begingroup$ @AndréNicolas I thought they were 2 conditions. $\endgroup$ – SalmonKiller Sep 2 '15 at 5:31

Here is how I would do it:

To satisfy a and b:

So the boy and the father should sit in adjacent chairs, so if the father sits on the left and the boy sits on the right, then there are 6 possible combinations. Flip the sides and you get 6 more: now we have 12. Now for each of the 8 ways such that there is a chair on either side, the mother can sit on either side of them. So we now have 16+4 = 20. For the other 4, there is only 1 chair for the mother to sit in, so the final answer is 20.

To satisfy a:

There are 12 ways we can sit the boy and the father as shown above. For each of those two ways, we have 5 ways to sit the mother. 12*5 = 60.

To satisfy b:

If we had 3 chairs, there would be 3!=6 ways to sit the three people. There are 5 ways to sit three people together if we have a row of 7 chairs. So 6*5 = 30 ways.

Ask in comments if you need more clarification.

  • $\begingroup$ You did not count mother in the middle. $\endgroup$ – André Nicolas Sep 2 '15 at 5:29
  • $\begingroup$ @AndréNicolas the father and the boy should sit together, so the mother can't sit in the middle. $\endgroup$ – SalmonKiller Sep 2 '15 at 5:30
  • 1
    $\begingroup$ I think of a) and b) as separate questions. But the other interpretation is not unreasonable. $\endgroup$ – André Nicolas Sep 2 '15 at 5:33
  • $\begingroup$ @AndréNicolas Thanks for pointing that out. I added the other solutions. $\endgroup$ – SalmonKiller Sep 2 '15 at 5:37

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