$4$ females and $6$ males will be seated on $19$ chairs

Question

There are $4$ females and $6$ males students. They will be seated on $19$ chairs. How many ways can we do this, if no two females are seated in adjacent chairs?

What makes me confused is there are only $10$ students but the chairs are $19$.

If the chairs are $10$ only (equal to the number of persons) I could do $C(7,4)\cdot 4!6!$

Answer 1

Here is one way of thinking it: First seat the men then seat the women. And arrange people in the right order before seating.

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\begin{align}P(15,6)\cdot P(16,4)\\=\boxed{{15\choose6}6!\cdot{16\choose4}4!}\end{align}

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It may sounds creepy, but here is another hint if you don't get it:

\begin{align}\textrm{Beyond those people, there are identical 9 ghost there(to help you)}\end{align}

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The key idea is about the $15$ and $16$:

• 15: this comes from 19-4, why? because I first consider those 6 men and 9 ghosts.
• 16: now we finish the queuing of 6+9=15 men-ghost-queue, what's the next step to make girls no adjacent?

Answer 2

Total Number of chairs=19.

Total number of persons=10.

So number of possible to select 10 chairs from 19 chairs $=19C10$.

Now take 10 chairs.

Then

$_B_B_B_B_B_B_$

If you're filling 4 girls in any 4 spaces among 7 spaces yiels a required arrangements. There are $7C4$ ways to choose filling 4 girls.

Thus totally we have $19C10×7C4×4!×6!$ ways.

Where $4!$ represents PERMUTATION among 4 girls and $6!$ represents PERMUTATION among 6 boys.

Answer 3

Let's first sit the females... For that first select $4$ chairs to them: ${19 \choose 4}$. Now we have left $19-4$ chairs remaining. Let's then distribute the remaining chairs between the females to help us with the restriction that no two females are seated in adjacent... For that, think as the females as separators, therefore: $$ \underbrace{ }_\text{slot 1}\text{ female 1}\underbrace{ }_\text{slot 2}\text{ female 2}\underbrace{ }_\text{slot 3}\text{ female 3}\underbrace{ }_\text{slot 4}\text{ female 4}\underbrace{ }_\text{slot 5} $$ where slots 2, 3 and 4 need to have at least one chair. From that we can create the following equation: $$ \underbrace{s_1}_{\geq 0 }+\underbrace{s_2}_{\geq 1 }+\underbrace{s_3}_{\geq 1}+\underbrace{s_4}_{\geq 1}+ \underbrace{s_5}_{\geq 0} = 19-4 = 15 $$ using stars and bars method we get that we can distribute the chairs in those slots in ${16 \choose 4}$ ways.

Now we've just sat 4 females and distributed the remaining $15$ chairs according to our restriction. Let's then choose $6$ of those $15$ chairs to sit the males. We can do that in ${15 \choose 6}$ ways.

Finally we have males and females sitting in a way that between two females we're going to have or chairs or males to separate them. We just need to organize them in all possible ways now: $4!6!$.

And our final answer, by multiplication principle is: $$ {19 \choose 4}{16 \choose 4}{15 \choose 6}4!6! $$

EDIT:

As Isana noticed in the comments, the ${19 \choose 4}$ at first is not needed since the distribution of chairs, at second step of the reasoning, will determine the position of the females. Therefore the corrected answer is: $$ {16 \choose 4}{15 \choose 6}4!6! $$