Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Arrangements of 0's, 1's, and 2's


Date: 03/14/2002 at 21:39:01
From: Sarah Moore
Subject: Arrangements of 0s, 1s and 2s

I have a question I hope you can help me with. How many arrangements 
of six 0's, five 1's, and four 2's are there in which 

  i) the first 0 precedes the first 1? 
 ii) the first 0 precedes the first 1, precedes the first 2?

For part i) I believe that there are C(15,2) ways to pick the two 
positions for the first 0 and first 1.  Then out of the remaining 13 
positions you can pick the positions for the four 2's in C(13,4) 
ways.  I am having difficulty figuring out how to continue.

Thanks again for your help.


Date: 03/15/2002 at 08:13:44
From: Doctor Anthony
Subject: Re: Arrangements of 0s, 1s and 2s

>  i) the first 0 precedes the first 1? 

This is more difficult than it first appears.  We have

   6 0's,   5 1's,    4 2's  =  15 digits

You must consider several non-overlapping configurations. For example, 
you could have 0 followed by every possible arrangement of the 
remaining digits, or 20 followed by every possible arrangement of the 
remaining digits, or 220 followed etc.

 First positions    Number and type of remaining digits
 --------------------------------------------------------
      0              5 0's, 5 1's, 4 2's  = 14!/[5!5!4!]
     20              5 0's, 5 1's, 3 2's  = 13!/[5!5!3!]
    220              5 0's, 5 1's, 2 2's  = 12!/[5!5!2!]
   2220              5 0's, 5 1's, 1 2    = 11!/[5!5!1!]
  22220              5 0's, 5 1's         = 10!/[5!5!0!]

            1     14!     13!     12!     11!      10! 
  Total = -----[ ----- + ----- + ----- + -----  + -----]
           5!5!    4!      3!      2!      1!       0!

        =  343980

>  ii) the first 0 precedes the first 1, precedes the first 2?

This time we MUST start with 0 and thereafter simply repeat the above 
calculation except that we now consider the digits 1 and 2 instead of 
0 and 1.  So having started with 0, we then have

    5 0's,   5 1's,   4 2's

I am now considering the arrangements of the 14 digits AFTER removing 
the first 0.


 First positions     Number and type of remaining digits
 --------------------------------------------------------
       1             5 0's, 4 1's, 4 2's = 13!/[5!4!4!]
      01             4 0's, 4 1's, 4 2's = 12!/[4!4!4!]
     001             3 0's, 4 1's, 4 2's = 11!/[3!4!4!]
    0001             2 0's, 4 1's, 4 2's = 10!/[2!4!4!]
   00001             1 0,   4 1's, 4 2's =  9!/[1!4!4!]
  000001             0 0's  4 1's, 4 2's =  8!/[0!4!4!]  

           1     13!    12!     11!     10!     9!      8!
 Total = -----[ ---- + ----- + ----- + ----- + ----- + ----]
          4!4!   5!     4!       3!      2!     1!      0!

       =  140140

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Permutations and Combinations

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/