2
$\begingroup$

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?

Thanks

$\endgroup$
  • 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.