How many groups of pentagonal flower bouquets can be formed?

Question

A florist has three types of flowers: tulips, roses, and daisies. There are 4 tulips, 5 roses, and 6 daisies. These 15 flowers are to be arranged into three bouquets of 5 flowers each. Assume that

• the order of the three bouquets is irrelevant,
• flowers of the same type are indistinguishable.

How many groups of pentagonal bouquets can the florist bundle?

Attempt

Let us denote tulips, roses, and daisies with T, R, and D, respectively. If we form all 15-letter strings and add dashes after every five letters, we can obtain all possible groups of bouquets. For instance, one possibility would be $$\mathrm{TRRTR-TRDDD-DDTRD}.\tag{ex. 1}$$

There are $\dfrac{15!}{4!\ 5!\ 6!}$ such strings. While, of course, all groups of bouquets can be obtained this way, we are overcounting. For strings, $\mathrm{TRDDD-TRRTR-DDTRD}$ is different from the example above, yet it makes no difference for the group of bouquets since order was assumed to be irrelevant. It might be tempting to divide the number of strings by $3!$ but this would also be incorrect. As an example, $\mathrm{TDDDT-TDDDT-RRRRR}\tag{ex. 2}$ is a valid group of three bouquets that should instead be divided by $\dfrac{3!}{2!} = 3$.

So, one way to proceed is to divide all groups of bouquets into two non-intersecting classes. First those for which all groups of three bouquets are pairwise different, and then those with exactly two matching bouquets out of three. Note that forming groups with three identical bouquets is impossible because 4 tulips cannot be shared equally among three bouquets. Once partitioned in this manner, we may divide the first kind of partition with $3!$, and the second with $3$.

However, such partitioning seems overly tedious, and is further complicated by the following aspect. We still have to consider that whenever there are at least two different kinds of flowers in a single bouquet, there is a further overcounting with the string-approach. Namely, e.g., the bouquets $$\mathrm{TRDDD\equiv DTRDD\equiv DDTRD\equiv DDDTR\equiv RDDDT}\tag{ex. 3}$$

are all equivalent since they can be transformed into one another by a rotation in space. (So a division with $5$ might additionally be in order for such bouquets). The 'further complication' is then the fact that groups of bouquets which initially seem to warrant division by $3!$, actually require division with $3$, as is the case for our first example. Indeed, by ex. 3 we have $\mathrm{TRDDD\equiv DDTRD}$ and so $$\mathrm{TRRTR-TRDDD-DDTRD\equiv TRRTR-TRDDD-TRDDD}$$ which should by divided by $3$.

Clarification from comments: bouquets which can be transformed into one another by reflection are not equivalent, and should be counted as different bouquets.

Question

The above discussion seems to lead into various subcases where mistakes could be easy to make, and is tedious to generalise. Is there cleaner approach? Regardless, an answer which carefully carries the above scheme to completion has value, too. For the record, the answer I obtain with the above method is $898$.

Edit: I have now also "confirmed" the answer $898$ with an independent Python program.

Trying to find partitions of the multiset $\{\mathrm{T}:4, \mathrm{N}:5, \mathrm{D}:6\}$ into classes of size five is something that I admittedly have not thought much about, but on the face of it would lead to undercounting, since, for instance, the multiset $\{\mathrm{D, D, R, R, T}\}$ would not differentiate between non-equivalent bouquets $\mathrm{DDRRT}$ and $\mathrm{DTDRR}$.

(This question is from the context of introductory combinatorics without recurrences, generating functions and so on).

Answer 1

Not an answer, but just a long comment, regarding the partition of the multiset you mention, thus when we don't consider the order of flowers in bouquets. However, the order of bouquets is taken into account here, so that e.g. TTDDD-TTDDD-RRRRR is different from TTDDD-RRRRR-TTDDD. I don't think this calculation can help to solve your problem, anyway.

We can define $t_1, t_2, t_3$ the number of tulips in bouquet $1,2,3$ respectively, $r_1, r_2, r_3$ the number of roses in bouquet $1,2,3$ respectively, $d_1, d_2, d_3$ the number of daisies in bouquet $1,2,3$ respectively. We need then to find the number of non-negative integer solutions to the following system of equations:

$$\begin{cases} t_1+r_1+d_1=5 \\ t_2+r_2+d_2=5 \\ t_3+r_3+d_3=5 \\ t_1+t_2+t_3=4 \\ r_1+r_2+r_3=5 \\ d_1+d_2+d_3=6 \end{cases}$$

Then we can use generating functions, and assign the following variables to each equation: $x$ to equation $1$, $y$ to equation $2$, $z$ to equation $3$, $t$ to equation $4$, $u$ to equation $5$, $v$ to equation $6$. The generating function is built with one factor for each variable of the linear system:

$$f(x,y,z,t,u,v)=\frac{x^6t^6-1}{xt-1}\frac{x^6u^6-1}{xu-1}\frac{x^6v^6-1}{xv-1}\frac{y^6t^6-1}{yt-1}\frac{y^6u^6-1}{yu-1}\frac{y^6v^6-1}{yv-1}\frac{z^6t^6-1}{zt-1}\frac{z^6u^6-1}{zu-1}\frac{z^6v^6-1}{zv-1} $$

For example, $\frac{x^6t^6-1}{xt-1}$ takes into account equation $1$ and $4$ for the variable $t_1$. We limit the terms to $x^5$ because this is the total of the first equation.

We use then WolframAlpha to compute the product of the first three factors, see here. We are interested only on the term with $x^5$ (the other factors don't contribute to $x$), which is:

$$x^5 (t^5 + t^4 (u + v) + t^3 (u^2 + u v + v^2) + t^2 (u^3 + u^2 v + u v^2 + v^3) + t (u^4 + u^3 v + u^2 v^2 + u v^3 + v^4) + u^5 + u^4 v + u^3 v^2 + u^2 v^3 + u v^4 + v^5)$$

and we can add the other two factors to get:

$$x^5 y^5 z^5 (t^5 + t^4 (u + v) + t^3 (u^2 + u v + v^2) + t^2 (u^3 + u^2 v + u v^2 + v^3) + t (u^4 + u^3 v + u^2 v^2 + u v^3 + v^4) + u^5 + u^4 v + u^3 v^2 + u^2 v^3 + u v^4 + v^5)^3$$

Then, we need to find the coefficient of $x^5y^5z^5t^4u^5v^6$. To do that, we still use WolframAlpha, but note that I had to replace $t,u,v$ with $x,y,z$ there only to make the engine understand the query, so this is a little confusing, but think of having $t,u,v$ there.

The result is $210$, I hope to have done all correctly. I don't know if there is a way to do this by hand in an elegant way.

I noticed that $\frac{15!}{4!5!6!}=630630=210 \times 3003$ but I don't know if this might mean something.

Answer 2

We begin by building flower arrangements ${\bf h}=(h_1,h_2,h_3)$, consisting of three heaps $h_i=(t_i,r_i,d_i)$ $(1\leq i\leq3)$, whereby the numbers $t_i$, $r_i$, $d_i$ are giving the numbers of tulips, roses, and daisies in heap $h_i$.

By stars and bars the $4$ tulips can be distributed among the $h_i$ in ${4+2\choose 2}=15$ ways, the $5$ roses in ${5+2\choose2}=21$ ways, and the $6$ daisies in ${6+2\choose2}-3=25$ ways (we don't want to put all $6$ daisies on the same heap). It follows that there are $15\cdot 21\cdot 25=7875$ ways to build such an arrangement. A little program produces all of these and checks for each whether it is admissible, i.e., all heaps have size $5$. This means that the additional conditions $$t_i+r_i+d_i=5\qquad(1\leq i\leq3)\tag{1}$$ are satisfied. It turns out that $210$ arrangements fulfill $(1)$.

An example is ${\bf h}=(212, 203, 041)$, where we have left out the inner commata. This ${\bf h}$ contains $2$ tulips, $1$ rose, and $2$ daisies in the first heap. Among the $210$ arrangements found by the program there is also ${\bf h}'=(203,041, 212)$ with the same three heaps, but in other order. Since the OP has desired that the heaps are not numbered we have to exclude such duplicates. Doing this leaves us with only $38$ arrangements, namely $$\eqalign{&(005, 041, 410), \quad (005, 050, 401),\quad (005, 131, 320), \quad(005, 140, 311), \quad(005, 221, 230),\cr &(014, 032, 410),\quad (014, 041, 401), \quad(014, 122, 320), \quad(014, 131, 311),\quad (014, 140, 302),\cr &(014, 212, 230), \quad(014, 221, 221)^*,\quad (023, 023, 410)^*,\quad (023, 032, 401), \quad(023, 113, 320),\cr & (023, 122, 311), \quad(023, 131, 302),\quad (023, 203, 230),\quad(023, 212, 221), \quad(032, 104, 320),\cr & (032, 113, 311),\quad (032, 122, 302), \quad(032, 203, 221),\quad (032, 212, 212)^*,\quad (041, 104, 311),\cr &(041, 113, 302), \quad(041, 203, 212), \quad(050, 104, 302), \quad(050, 203, 203)^*, \quad(104, 122, 230),\cr &(104, 131, 221),\quad (104, 140, 212), \quad(113, 113, 230)^*,\quad (113, 122, 221), \quad(113, 131, 212),\cr & (113, 140, 203), \quad(122, 122, 212)^*,\quad (122, 131, 203).\cr}$$$6$ of these arrangements, marked with an asterisk, have two equal heaps. They need some special treatment in what follows.

So far the $h_i$ were just heaps of five flowers. But the OP wants to create regular pentagons from these flowers, whereby rotations of the same pentagon should be counted only once. The number triples occurring in the above list are permutations of the five triples $$500,\quad 410,\quad320,\quad 311,\quad 221\ .$$ The three appearing numbers give the numbers of flowers of each color in the heap. Each of these triples $h$ allows a certain number $m(h)$ of rotationally different pentagons. $500$ means that all five flowers have the same color. In this case there is only one pentagon possible, and the same is true for $410$; hence $m(500)=m(410)=1$. With $320$ the two equal flowers can have a distance of $1$ or $2$ in the pentagon; makes $m(320)=2$. Then $311$: The first single flower can be placed anywhere, the second single flower at four places, makes $m(311)=4$. Finally $221$: The single flower can be placed anywhere, and the first pair of equals can take its places in ${4\choose2}$ ways; makes $m(221)=6$.

One now has to go through the $38$ arrangements in the above list, and to multiply the $m$-values of the three heaps, when there are three different heaps. In the starrred arrangements it's a bit more complicated: When ${\bf h}=(h,h,h')$ then we can display the two $h$ either differently in ${m(h)\choose2}$ ways or equally in $m(h)$ ways. The total number of possible displays for this ${\bf h}$ therefore is $$\left({m(h)\choose2}+m(h)\right)\cdot m(h')\ .$$

Sum all the obtained products (or corrected values) up, and you have the final result.