2
$\begingroup$

There are $4$ girls and $3$ boys but there are only $5$ seats. How many ways can you seat the $3$ boys together?

The order of the seat matters, for example: there's the order $B_1$ $B_2$ $B_3$ $G_2$ $G_4$ and there's $B_2$ $B_3$ $B_1$ $G_2$ $G_4$

Here's my answer: There are $3!$ ways to seat the $3$ boys. The $2$ remaining seats are to be occupied by $2$ out of the $4$ girls, so $^4P_2$. So we now have $3! \cdot $ $^4P_2$.

Lastly, there are $3$ ways to make that arrangement,
$1)$ two girls on the left,
$2)$ two girls on the right, and
$3)$ a girl on both ends.

So my final equation is $3! \cdot $ $^4P_2$ $\cdot 3 = 216$

But then again, that was just a guess, I'm not really sure how to get it. So please confirm if my answer is right, and if it's wrong, please tell me how to get it.

$\endgroup$
1
  • 1
    $\begingroup$ Seems correct to me. $\endgroup$
    – user35359
    Oct 2, 2015 at 3:26

2 Answers 2

1
$\begingroup$

You are correct. Here is another approach:

We can select the two girls in $\binom{4}{2}$ ways. We treat the three boys as a unit, so we have three objects to permute (the two girls we select and the unit of three boys). We can permute the three objects in $3!$ ways. We can also permute the unit consisting of three boys internally in $3!$ ways. Hence, there are $$\binom{4}{2} \cdot 3! \cdot 3! = 6^3 = 216$$ seating arrangements in which the seats are occupied by the three boys and two of the four girls if the three boys sit together.

$\endgroup$
0
$\begingroup$

First Take all of the boys in a group say $A$. $A$ contains $B_1$, $B_2$ and $B_3$. Then select two girls from $4 \Rightarrow$ $^4C_2$. Now we got $2$ girls and boys (in $A$) which makes it $5$. We can arrange $B_1$ $B_2$ $B_3$ inside $A$ in $3!$ ways . and we can arrange $A$ and the $2$ girls in $3!$ ways . Hence the answer would be $^4C_2 \cdot 3! \cdot 3! =216.$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .