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
Attempt:
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?
Best Answer
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 $4!*4!$