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### 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.

```

```
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

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