All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Probability
TexMex Take-out (Posted on 2007-02-25) Difficulty: 3 of 5
A restaurant offers a take-out deal where any combination of 10 items can be purchased for $5. The menu choices are divided into five types: Tacos, Burritos, Tamales, Chiles Rellenos, and Chimichangas. For example, a person could order 2 Tacos, 4 Burritos, a Tamal and 3 Chimichangas. Or maybe 7 Tamales and 3 Chiles Rellenos. Or perhaps 10 Chimichangas and nothing else. How many different ten-item combinations are available?

No Solution Yet Submitted by hoodat    
Rating: 4.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution TexMex Nights (spoiler) Comment 2 of 2 |
Think about 14 items, 4 of which are separators (X's) between types and the 10 of which are actual items (O's).  The items are always ordered with Tacos first, then Burritos, then Tamales, then Chille Rellenos, then Chimichangas.

In this scheme, 2 Tacos, 4 Burritos, a Tamale, and 3 Chimichangas would be represented as OOXOOOOXOXOOO

7 tamales and 3 Chile Rellenos would be represented as: XXOOOOOOOXOOOX

10 Chimichangas would correspond to: XXXXOOOOOOOOOO

So this problem is equivalent to determining all the different arrangement of 4 seperators among 14 spaces. 

This equals C(14,4) = 14!  /  4!  * 10 ! =
14*13*12*11/(4*3*2*1) = 1001



  Posted by Steve Herman on 2007-02-25 16:16:16
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information