# Help with a (non-homework) math problem **URL:** \ **Category:** Factual Questions\ **Created:** [October 3, 2011, 4:32pm UTC](https://boards.straightdope.com/t/help-with-a-non-homework-math-problem/598446 "2011-10-03T16:32:37Z")\ **Posts on this page:** 11\ **Page:** 1
**Author:** ![Elret](https://avatars.discourse-cdn.com/v4/letter/e/b9bd4f/32.png) [@Elret](https://boards.straightdope.com/u/Elret)\ **Post date:** [October 3, 2011, 4:32pm UTC](https://boards.straightdope.com/t/help-with-a-non-homework-math-problem/598446/1 "2011-10-03T16:32:37Z")
I am sure I learned this back in the day, but can’t figure it out. What would be the steps I would take to solve the following math problem? You are going to make some teddybears. Each bear will have a hat, a ribbon, and a vest. You have 7 different colours of material for the hats, ribbons, and vests. No two bears will be wearing the exact same “outfit”, and no one bear will be wearing two or more things of the same colour. How many bears would you need to make so that you use up every possible colour combination? ---
**Author:** ![Buck\_Godot](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/buck_godot/32/6573_2.png) [@Buck\_Godot](https://boards.straightdope.com/u/Buck_Godot)\ **Post date:** [October 3, 2011, 4:36pm UTC](https://boards.straightdope.com/t/help-with-a-non-homework-math-problem/598446/2 "2011-10-03T16:36:32Z")
In general there are 7^3=343 combinations of the 7 colors for three accessories. 7 of these are sets of all the same colors (one for each color). So there are 343-7= 336 possible combinations that don’t have the same color. I assume that red hat, red ribbon, blue vest is treated as a separate combination from red hat, blue ribbon, red vest. ---
**Author:** ![Elret](https://avatars.discourse-cdn.com/v4/letter/e/b9bd4f/32.png) [@Elret](https://boards.straightdope.com/u/Elret)\ **Post date:** [October 3, 2011, 4:39pm UTC](https://boards.straightdope.com/t/help-with-a-non-homework-math-problem/598446/3 "2011-10-03T16:39:19Z")
Thanks! Can you walk a dummy through how to come to that? Oh, and each bear must be wearing three different colours. ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [October 3, 2011, 4:40pm UTC](https://boards.straightdope.com/t/help-with-a-non-homework-math-problem/598446/4 "2011-10-03T16:40:23Z")
> [@Elret](#): > > I am sure I learned this back in the day, but can’t figure it out. What would be the steps I would take to solve the following math problem? > > You are going to make some teddybears. Each bear will have a hat, a ribbon, and a vest. You have 7 different colours of material for the hats, ribbons, and vests. No two bears will be wearing the exact same “outfit”, and no one bear will be wearing two or more things of the same colour. How many bears would you need to make so that you use up every possible colour combination? The simple answer is: 7_6_5=210 (Because for each of the seven hat colors, there are six possible vest colors (since the two can’t be the same color) and for each of those possible vest colors, there are five possible ribbon colors.) ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [October 3, 2011, 4:41pm UTC](https://boards.straightdope.com/t/help-with-a-non-homework-math-problem/598446/5 "2011-10-03T16:41:25Z")
> [@Buck\_Godot](#): > > In general there are 7^3=343 combinations of the 7 colors for three accessories. 7 of these are sets of all the same colors (one for each color). So there are 343-7= 336 possible combinations that don’t have the same color. > > I assume that red hat, red ribbon, blue vest is treated as a separate combination from red hat, blue ribbon, red vest. This allows a bear to wear two things of the same color–the OP says that’s not allowed. ---
**Author:** ![ENugent](https://avatars.discourse-cdn.com/v4/letter/e/4af34b/32.png) [@ENugent](https://boards.straightdope.com/u/ENugent)\ **Post date:** [October 3, 2011, 4:42pm UTC](https://boards.straightdope.com/t/help-with-a-non-homework-math-problem/598446/6 "2011-10-03T16:42:37Z")
There are seven possible colors for the first bear’s hat. Once the hat color is chosen, there are six colors left that are possible choices for his vest (since it can’t be the same color as the hat). Once the vest color is chosen, there are five possible choices for his ribbon. So there are 7x6x5 = 210 possible combinations. ---
**Author:** ![Asympotically\_fat](https://avatars.discourse-cdn.com/v4/letter/a/e47c2d/32.png) [@Asympotically\_fat](https://boards.straightdope.com/u/Asympotically_fat)\ **Post date:** [October 3, 2011, 4:43pm UTC](https://boards.straightdope.com/t/help-with-a-non-homework-math-problem/598446/7 "2011-10-03T16:43:48Z")
we can divide the bears in to 7 groups based on their hat colour We can further divided each of these groups in to 6 groups based on their ribbon colour We can yet again divide each of the 2nd groups into 5 groups based on their vest colour. As each bear is wearing a unique outfit each of these 3rd groups will consist of 1 bear each. So total number of bears = 5_6_7 = 210 bears ---
**Author:** ![Elret](https://avatars.discourse-cdn.com/v4/letter/e/b9bd4f/32.png) [@Elret](https://boards.straightdope.com/u/Elret)\ **Post date:** [October 3, 2011, 4:49pm UTC](https://boards.straightdope.com/t/help-with-a-non-homework-math-problem/598446/8 "2011-10-03T16:49:51Z")
Ahhh. That makes perfect sense. Thanks, all! ---
**Author:** ![Buck\_Godot](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/buck_godot/32/6573_2.png) [@Buck\_Godot](https://boards.straightdope.com/u/Buck_Godot)\ **Post date:** [October 3, 2011, 5:39pm UTC](https://boards.straightdope.com/t/help-with-a-non-homework-math-problem/598446/9 "2011-10-03T17:39:44Z")
> [@Frylock](#): > > This allows a bear to wear two things of the same color–the OP says that’s not allowed. Sorry I misread the OP. I agree with the 7_6_5. ---
**Author:** ![Chessic\_Sense](https://avatars.discourse-cdn.com/v4/letter/c/7c8e57/32.png) [@Chessic\_Sense](https://boards.straightdope.com/u/Chessic_Sense)\ **Post date:** [October 3, 2011, 6:01pm UTC](https://boards.straightdope.com/t/help-with-a-non-homework-math-problem/598446/10 "2011-10-03T18:01:38Z")
Look up [Rule of Product](http://en.wikipedia.org/wiki/Rule_of_product), or what I learned as the “multiplication counting principle”. It’s the root of all permutation and combination rules. It says that if there are a ways to make the first choice and b ways to make the second choice…z ways to make the 26th choice, the possibilities total to a_b_…\*z. I’ll try to make this intuitive. Imagine if you tried to write out all the combinations. Suppose I said to you “There are just seven hat colors. Write all the combinations out.” You’d have no problem with that. You’d just write: ```auto R O Y G B I V ``` You’d see that you need 7 bears. But then I come up to you and say “Hey, we’ve just decided we have our bears to have vests. There are still seven colors, but they can’t be the same as the hat the bears already picked.” Well, you’d look at your single red bear and say “I’ll need more bears, so that I can put six vests on red-hatted bears.” So then you’d have: ```auto RO, RY, RG, RB, RI, RV O Y G B I V ``` Hey, look. The single red bear became six unique bears. It’s almost like your stuffed animals are multiplying (see what I did there?) And then you’d guess that you’d do the same thing with the other bears. In fact, each single bear would become six bears. Hmm…7 bears, each individual one ‘multiplies’ and becomes 6…that sounds an awful lot like 7\*6 = 42. So now you’ve got 42 unique bears sitting on your dressing table and I come up and say “Y’know, we like the hats, we like the vests, but now we want ribbons. Same colors, same rules as the vests. Kthxbai.” Now you’d have to figure out how many bears you’ll need. So you take a single bear from the pile like the RO bear. ```auto ROY, ROG, ROB, ROI, ROV RY RG RB RI RV ...(and 36 others) ``` That single RO bear became five bears. It’s like it had babies…like it multiplied. By five. And you started with 42 bears. So each bear will become five bears. 42\*5 = 210. And that’s why the multiplication counting principle works. With every choice, every unique thing you already had multiplies by the amount of new choices there are. ---
**Author:** ![Elret](https://avatars.discourse-cdn.com/v4/letter/e/b9bd4f/32.png) [@Elret](https://boards.straightdope.com/u/Elret)\ **Post date:** [October 3, 2011, 8:49pm UTC](https://boards.straightdope.com/t/help-with-a-non-homework-math-problem/598446/11 "2011-10-03T20:49:35Z")
Thank you, that makes a lot of sense.