In how many ways can three boys and four girls occupy seven seats in a row if

a. A girl and a boy occupy the end seats

b. If the four girls must sit together


For the part a

The probability that a boy and a girl occupy the end seats. The boy and the girl can seat in $2!$ ways, and the other $5$ people can occupy the rest seat in $5!$ ways ... $$= 2! \times 5! = 2 \times 1 \times 5 \times 4 \times 3 \times 2 \times 1 = 240$$ ways.

This is what I think, I'm not sure

Part b

This is the way I think it should be

If the four girls must sit together.. We can first ask the boys to seat together and there are $3!$ possible ways. Since the $4$ girls must sit together, they have the following four choices for their positions. GGGGBBB or BGGGGBB or BBGGGGB or BBBGGGG Where (B) denote Boy and (G) denote Girl. Therefore, their are total of $3! \times 4 \times 4!$. So the number of ways the $4$ girls can seat together is $$= 3! \times 4 \times 4! = 3 \times 2 \times 1 \times 4 \times 4 \times 3 \times 2 \times 1 = 576$$ ways.

For this part b, someone told me the answer is $6$ but I got $576$. So please am I doing anything wrong?

  • $\begingroup$ You have to decide whether the treat ... Ann Belinda ... as being the same way as ... Belinda Ann ... . In your attempt at part (a) you say there are just 2 ways of filling the end seats, ie B ... G and G... B. That assumes we treat all boys as identical. But then you say the other 5 can be placed in 5! ways. That assumes they are not identical. $\endgroup$
    – almagest
    Commented Sep 26, 2015 at 10:45
  • $\begingroup$ Uh. The question says a boy and a girl. That's why I just chose B and G. $\endgroup$
    – user274246
    Commented Sep 26, 2015 at 10:56
  • $\begingroup$ I suspect it just wants arrangements of B, G too. So do you really think ther e are 120 ways of filling the middle 5 seats in part (a)? Try listing them. $\endgroup$
    – almagest
    Commented Sep 26, 2015 at 10:59
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. Note that you are not calculating a probability in part (a). What you are calculating is the number of ways the boys and girls may be seated. $\endgroup$ Commented Sep 26, 2015 at 11:27
  • $\begingroup$ Oh. I'm Sorry. I'm using an android. That's why I just typed it. $\endgroup$
    – user274246
    Commented Sep 26, 2015 at 11:40

2 Answers 2


Unless otherwise specified,

I'd take each girl and boy as distinct. After all, we aren't talking of apples and oranges.

(a) $2$ choices of ends for girl/boy.

$4*3 = 12$ ways to fill the ends with particular girl/boy

$5!$ ways to permute the rest,

thus $2*12*5! = 2880$

(b) Your ans is correct, but a simpler way is to treat the 4 girls as an internally permutable block $[GGGG]$, and permute the $4$ entities, $[GGGG]BBB$, thus $4!*4!$

  • $\begingroup$ I was in the process of writing this when I saw your answer. $\endgroup$ Commented Sep 26, 2015 at 11:36
  • $\begingroup$ I just realized that while you explained part (a) correctly, you forgot to multiply by $2$. $\endgroup$ Commented Sep 26, 2015 at 11:44
  • $\begingroup$ You mean multiply by two so the 12 ways will be for each end right? $\endgroup$
    – user274246
    Commented Sep 26, 2015 at 11:53
  • $\begingroup$ @user274246 That is correct. $\endgroup$ Commented Sep 26, 2015 at 11:56
  • $\begingroup$ Thanks. @Mr N.F Taussig. Please I use an android mobile. It doesn't support mathjax. How can I type then? $\endgroup$
    – user274246
    Commented Sep 26, 2015 at 12:06

It's not really clear from the question if internal position matters. Are we only interested in the sex and not the person for each position?

For b) You can get 6 if you think 4 specific girls sit at a specific position and does not matter in which order the girls sit (within the group), but it matters which orders the boys sit, then it would be 3 positions for first boy, 2 for second and 1 for last = $3\cdot 2 \cdot 1 = 6$. Otherwise if both internal position of girls and boys matters and placement of the group of girls, then it should be $3!\cdot 4! \cdot 4 = 576$ which you got.

So the question is.., what is the question giver really interested in knowing? How we choose to interpret the question maybe...

  • $\begingroup$ The order in which the individuals sit matters. See true blue anil's answer. $\endgroup$ Commented Sep 26, 2015 at 11:41
  • $\begingroup$ Oh thanks. I didn't look at it from your first perspective. That's if the girls sit at a specific position. Now I understand. $\endgroup$
    – user274246
    Commented Sep 26, 2015 at 11:44
  • $\begingroup$ @N.F.Taussig On balance, I don't agree with you. I think the question is just after BG... etc (see comments under question). $\endgroup$
    – almagest
    Commented Sep 26, 2015 at 12:10
  • $\begingroup$ @almagest In my experience, questions about arranging people treat each person as an individual. If your interpretation were intended, the question would have been about red and blue marbles. $\endgroup$ Commented Sep 26, 2015 at 12:16

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