In how many ways can $5$ boys and $5$ girls stand in a queue such that exactly four of the girls stand consecutively in the queue?

Question

I have two combinatorics questions.

Let $n$ be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let $m$ be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue. Then, the value of $m/n$ is?

Let $n_1<n_2<n_3<n_4<n_5$ be positive integers such that $n_1+n_2+n_3+n_4+n_5 = 20$. The number of distinct arrangements of ($n_1, n_2, n_3, n_4, n_5$) is?

So for the first question, I was able to find out $ n = 6! × 5!$ but while finding $m$, I could do $4! × 7!$ But that also include all the fives girls accidentally coming together.. So how do I eliminate those cases? I can't proceed further.

The second one, I thought was one of those stars and bars problems, so I did $C(24, 4)$ but then I realized that the condition on the values of numbers is not that simple. I tried converting it into simpler problem, as they do with stars and bars problems but I couldn't achieve something useful. I'm stuck :/

Can you help me get through these questions?

P.S.-These questions are sometimes meant to be done using a trick, so if you think you know some trick to make that easier, please be sure to tell it. And otherwise a true solution would be as helpful :)

Answer 1

In how many ways can $5$ boys and $5$ girls stand in a queue if all five girls stand consecutively in the queue?

You are correct that there are $6!5!$ ways for all five girls to stand consecutively in the queue.

Method 1: We treat the block of five girls as a single object. We then have six objects to arrange, the block of girls and the five boys. The objects can be arranged in $6!$ ways. The five girls can be arranged within the block in $5!$ ways. Thus, there are $6!5!$ ways for five boys and five girls to stand in a queue if all five girls stand consecutively in the queue.

Method 2: Line up the five boys, which can be done in $5!$ ways. This creates six spaces in which to place the block of five girls, four between successive boys and two at the ends of the row. $$\square b_1 \square b_2 \square b_3 \square b_4 \square b_5 \square$$ Choose one of these six spaces in which to place the block of girls, then arrange the five girls within the block. This can be done in $6 \cdot 5!$ ways. Hence, the number of admissible arrangements is $6!5!$.

In how many ways can $5$ boys and $5$ girls stand in a queue if exactly four girls stand consecutively in the queue?

We modify the second method above.

Line up the five boys in $5!$ ways. This creates six spaces in which to place the girls. Choose which four of the five girls stand consecutively, which can be done in $\binom{5}{4}$ ways. Choose which of the six spaces the block of four girls fills. Arrange the four girls in that space in $4!$ ways. That leaves five spaces in which to place the remaining girl. Hence, the number of ways five boys and five girls can stand in a queue if exactly four girls stand consecutively is $$5!\binom{5}{4} 6 \cdot 4! \cdot 5 = 5! \cdot 5 \cdot 6 \cdot 5 \cdot 4! = 5 \cdot 6!5!$$

In how many ways can $20$ be expressed as the sum of five distinct increasing positive integers?

Since $20$ is a small number, we can simply write down all the possibilities: \begin{align*} 20 & = 1 + 2 + 3 + 4 + 10\\ & = 1 + 2 + 3 + 5 + 9\\ & = 1 + 2 + 3 + 6 + 8\\ & = 1 + 2 + 4 + 5 + 8\\ & = 1 + 2 + 4 + 6 + 7\\ & = 1 + 3 + 4 + 5 + 7\\ & = 2 + 3 + 4 + 5 + 6 \end{align*} Notice that any sum of five distinct positive integers is at least $1 + 2 + 3 + 4 + 5 = 15$. We then have to distribute five more ones in such a way that we preserve the increasing sequence. Since $5$ can be partitioned into at most five positive integers in the following seven ways, \begin{align*} 5 & = 5\\ & = 4 + 1\\ & = 3 + 2\\ & = 3 + 1 + 1\\ & = 2 + 2 + 1\\ & = 2 + 1 + 1 + 1\\ & = 1 + 1 + 1 + 1 + 1 \end{align*} we can do so in the following ways: \begin{align*} (0, 0, 0, 0, 5)\\ (0, 0, 0, 1, 4)\\ (0, 0, 0, 2, 3)\\ (0, 0, 1, 1, 3)\\ (0, 0, 1, 2, 2)\\ (0, 1, 1, 1, 2)\\ (1, 1, 1, 1, 1)\\ \end{align*} Adding these, respectively, to the vector $(1, 2, 3, 4, 5)$ yields the solutions \begin{align*} (1, 2, 3, 4, 10)\\ (1, 2, 3, 5, 9)\\ (1, 2, 3, 6, 8)\\ (1, 2, 4, 5, 8)\\ (1, 2, 4, 6, 7)\\ (1, 3, 4, 5, 7)\\ (2, 3, 4, 5, 6) \end{align*} that correspond to the seven sums we wrote above.

Answer 2

Q1 solution:

No of ways all $5$ girls stand in a queue is obtained by considering all the $5$ girls as a single entity and then permute them along with $5$ boys . So, $n$ simply comes as: $6!×5!$ (where the later $5!$ is the no of ways to permute all the girls among themselves). Now, let's find out the no of ways $4$ girls can stand in a queue. First we choose $4$ girls from $5$ girls in ${5\choose 4} =5$ ways. Now as before, consider these $4$ girls as a single entity and considering their permutations along with remaining one girl and $5$ boys ,we get: $5×7!×4!$. But,in these cases,there are $2n$ cases where all the five girls are adjacent to each other. To see this, consider a permutaion like this- >$$G_1:G_2:G_3:G_4:G_5:B_1:...:B_5$$

This can happen when you have chosen the 4 girls(those that you are cosidering as a single entity) as $\{G_1,G_2,G_3,G_4\}$ or as $\{G_2,G_3,G_4,G_5\}$. So,there are exactly two cases of each such permutation. In particular, $m$ contains exactly $2n$ no of cases with $5$ girls in a row.

So $$m= 5×7!×4!-2.n$$