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.

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

2 Answers 2


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.


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 $^4C_2 \cdot 3! \cdot 3! =216.$


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