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I need to find out the number of possible combinations for the following, and would like to understand the working behind it as well.

I have a product that consists of 3 separate parts, head, arms and legs.

The product must always have the 3 parts present and must always be in the order head > arms > legs.

If I have 2 products, I could swap the heads for example. Then maybe swap the legs.

So, given these parameters, how many different combinations could I get if I have a collection of 6 products?

Mr Pablo
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2 Answers2

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Look at it as $6$ teams of $3$ being formed.

The $6$ "heads" are allowed to choose one each of a set of "arms" and "legs"

The $1st$ head can select its team in $6\cdot6 = 6^2$ ways.

Now only $5$ sets of arms and legs are left to choose from for the $2nd$ head,
who has $5\cdot5 = 5^2$ choices, and so on.

Thus # of different teams possible $= (6\cdot 5\cdot4\cdot3\cdot2\cdot1)^2 = (6!)^2 = 518400$

PS
Note that the above counts the number of divisions into $6$ products using all the parts. If each product in such a division is to be counted, you need to multiply by $6$

true blue anil
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  • Why isn't it (6!)^3? As there are 3 pieces? Also remember that a product has to be in order, head then arms then legs. – Mr Pablo Dec 04 '15 at 18:42
  • For any given "head" $(H)$, we can only choose $A$ and $L$ , so (6!)^2. Also, since we are only choosing $H-A-L,$ we can always assemble them in correct order. Why don't you enumerate all possibilities for two sets each of $H,A,L$ – true blue anil Dec 05 '15 at 03:50
  • I did (6!)^2 and I get an answer of 518,400 – Mr Pablo Dec 05 '15 at 10:41
  • My bad ! Typo corrected ! – true blue anil Dec 05 '15 at 11:45
  • Surely the fact you are choosing a head first means it's to the power of 3,as there are 3 parts? – Mr Pablo Dec 05 '15 at 14:52
  • Take the case of $2$ each of $H,A,L$. Withh $H1$, you can only make $H_1A_1L_1, ;H_1A_1L_2,; H_1A_2L_1,; H_1A_2L_2$ so there will be only $2^2 = 4$ sets with each type of head (i.e. 4 divisions into teams in my solution) If you want to count each team $=$ each product, for the above case you need to multiply by $2$, and for the $6$ case, multiply by $6$ – true blue anil Dec 05 '15 at 15:07
  • I don't quite understand your PS edit. What do I need to multiply by 6? – Mr Pablo Dec 05 '15 at 23:39
  • For the $2$ case, each head can give rise to $4$ products, so $4\times2$ total products. But if you put all you can make using all the parts, and put them in a display window, there can only be $4$ at a time. Same thing for $6$. There can be $518,400$ displays at a time, but $6$ times as many distinct products possible. – true blue anil Dec 06 '15 at 04:18
  • Still not sure what you mean by "but 6 times as many distinct products possbile"? – Mr Pablo Jan 05 '16 at 14:10
  • For the two heads case, if you want to display in a window using all parts at any one time, you have 4 sets: H1A1L1 - H2A2L2 or H1A1L2 - H2A2L1 or H1A2L1 - H2A1L2 or H1A2L2 - H2A1L1, but there are 8 products i.e twice as many products, because each of 2 heads forms a product. Similarly for 6 heads. – true blue anil Jan 05 '16 at 14:34
  • I've just come back to this for some clarity, and still don't understand the last part I'm afraid. Let's stick with the example of having 6 separate, 3-piece characters. Say I want to display 6 characters in a window, each day changing the combination to one of the unique combinations. Is that just (6!)^2 ? – Mr Pablo Dec 06 '16 at 15:59
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If you have $6$ products, then you have $6$ heads, $6$ arms, and $6$ legs. Building a new product from these parts is picking $1$ head, $1$ arm, and $1$ leg from each of the piles of $6$, so you have $6\cdot6\cdot6=6^3$ possible products.

Ryan Vitale
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