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  • There are $3$ boys and $3$ girls. How many ways can they be arranged in a way that two girls won't be together?

Instead of making things harder, I directly write an assumption as illustrated below

$$\text{All permutations = Total possibilities - Two girls are together}$$

We already notice that total possibilities should be $6!$. Let's calculate the permutation of two girls are together. Recall that $K$ represents Girls, $E$ represents boys.

Entity $1$: $K_1K_2 \implies 1$

Entity $2$: $B_1B_2B_3K_3 \implies 4$

Entity $1$ and $2$: $5!$

And girls can be rearranged in $2$, which yields $5!\times 2!$. Plugging into the equation we wrote

$$6!-5!\times 2! = 480$$

However, the final answer I got should be wrong. Can you assist?

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  • $\begingroup$ You could have two girls together or you could have all three together. $\endgroup$ – Neal Oct 11 '18 at 19:03
  • $\begingroup$ @Neal Is my assumption wrong? If yes, why? $\endgroup$ – Mark Oct 11 '18 at 19:06
  • $\begingroup$ The complement of "two girls not together" is "at least two girls together". So you need to count two girls together. You also need to count the ways you can have all three girls together. $\endgroup$ – Neal Oct 11 '18 at 19:11
  • $\begingroup$ You meant the total number of possibilities is $6!$, not the total probability. $\endgroup$ – N. F. Taussig Oct 11 '18 at 19:25
  • $\begingroup$ @N.F.Taussig Could you be more specific? I cannot properly obtain where the issue is. $\endgroup$ – Mark Oct 11 '18 at 19:26
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All the ways to seat all of the people

$6!$

At least $2$ girls together.

We attach 2 girls, and allocate $5$ objects. Making for $5!$ allocations

$5!$ But we can swap the order of the girls. And we have 3 girls who we can make the "odd-man-out"

$5!\cdot 6 = 6!$

But we have double counted all of the cases where we have all 3 girls together.

$4!3!$

$6! - (6! - 4!3!) = 4!3! = 144$

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You don't state it, but I'm assuming the boys are indistinguishable and the girls are indistinguishable. Please clarify this point!

Consider three girls. There must be at least one boy separating pairs: $GBGBG$. The only question then remains where the last boy can go: $\bullet G \bullet B \bullet G \bullet B \bullet G \bullet$. Note that some of these cases are identical, so you can eliminate them

How many distinguishable places all told?

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  • $\begingroup$ I dont know how to calculate that. And is my assumption wrong? $\endgroup$ – Mark Oct 11 '18 at 19:05
  • $\begingroup$ People are considered to be distinguishable unless otherwise stated. $\endgroup$ – N. F. Taussig Oct 11 '18 at 19:55
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In how many ways can three boys and three girls be arranged in a row so that no two girls are adjacent?

Method 1: We arrange the three boys in a row, then place the girls in the spaces between the boys and at the ends of the row to separate them.

The three boys can be arranged in a row in $3!$ ways. This creates four spaces, two between successive boys and two at the ends of the row. $$\square B_1 \square B_2 \square B_3 \square$$ To ensure that no two girls are adjacent, we must choose of these spaces in which to place the girls, which can be done in $\binom{4}{3}$ ways. The three girls can be arranged in the selected spaces in $3!$ ways. Hence, the number of admissible arrangements is $$\binom{4}{3}3!3!$$

Method 2: We correct your attempt by using the Inclusion-Exclusion Principle.

There are $6!$ ways to arrange the six people. As you stated, we wanted to exclude those arrangements in which a pair of girls are adjacent.

There are $\binom{3}{2}$ ways to select two girls who are adjacent. We have five objects to arrange, the block of two girls and the other four children. The objects can be arranged in $5!$ ways and the girls can be arranged within the block in $2!$ ways. Hence, there are $$\binom{3}{2}5!2!$$ arrangements in which two girls are adjacent.

If we subtract the number of arrangements in a pair of girls are adjacent from the total, we will have subtracted too much since we will have subtracted each arrangement in which there are two pairs of adjacent girls twice, once for each pair of adjacent girls we could have designated as the pair of adjacent girls. We only want to subtract those arrangements once, so we have to add the number of arrangements in which there are two pairs of adjacent girls to the total.

To have two pairs of adjacent girls, all three girls must be consecutive. If we treat the three consecutive girls as a block, we have four objects to arrange, the block of three girls and the three boys. The objects can be arranged in $4!$ ways. The girls can be arranged within the block in $3!$ ways. Hence, the number of arrangements in which there are two pairs of adjacent girls is $4!3!$.

Hence, the number of admissible arrangements is $$6! - \binom{3}{2}5!2! + \binom{3}{3}4!3!$$

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The girls can be in positions $\{1,3,5\}$, $\{1,3,6\}$, $\{1,4,6\}$, or $\{2,4,6\}$. For each of these there are six permutations of the girls, and for each of these there are also six permutations of the boys, so the answer should be $4\cdot 3!\cdot 3! = 144$.

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    $\begingroup$ The logic is fine, but $4*6*6 = 144$ $\endgroup$ – Doug M Oct 11 '18 at 19:40
  • $\begingroup$ @DougM What's wrong with $6-5! \times 2! + 4! \times 3!$? I counted the ways I can have all girls together, which equals $4! \times 3!$ and the ways I can have two girls together, $5! \times 2!$ $\endgroup$ – Mark Oct 11 '18 at 19:41

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