Quick Elementary Combinatorics Question / Verification

Click For Summary
SUMMARY

The discussion revolves around calculating the number of valid variable names in a programming language based on specific rules. For exactly 5 characters, there are 834,906,933 possible variable names, calculated using 53 options for the first character and 63 options for the remaining four. For variable names with at most 5 characters, the total is 848,373,173. When considering palindromic names with at most 5 characters, the total is 217,141 valid names. These calculations are confirmed as accurate by other participants in the discussion.

PREREQUISITES
  • Understanding of combinatorial mathematics
  • Familiarity with variable naming conventions in programming
  • Basic knowledge of character sets and their applications
  • Ability to perform summation and multiplication of large numbers
NEXT STEPS
  • Explore combinatorial counting techniques in programming languages
  • Learn about character encoding and its impact on variable naming
  • Study palindromic sequences and their applications in algorithms
  • Investigate variable naming conventions across different programming languages
USEFUL FOR

Students in computer science, programmers interested in variable naming rules, and educators teaching combinatorial mathematics.

tylerc1991
Messages
158
Reaction score
0

Homework Statement



I am using a book that doesn't have any solutions in it, so I would like to be sure that I am doing the problems right before I move on. The question is below:

In a programming language, a variable name must start with a letter or the underscore character, and succeeding characters must be letters, digits, or the underscore character. Uppercase and lowercase letters are considered different.
(a) How many variable names with exactly 5 characters can be formed?
(b) How many are there with at most 5 characters?
(c) How many are there with at most 5 characters, if they must read the same forwards and backwards? (note that kayak is acceptable but Kayak is not).

The Attempt at a Solution


(a) There are 53 possible characters for the first position, and 63 possibilities for the remaining 4 positions. So there are
[itex]53 \cdot 63^4 = 834,906,933[/itex]
possible 5 character variable names.

(b) For some [itex]k[/itex], there are [itex]53 \cdot 63^{k-1}[/itex] possible variable names of length [itex]k[/itex]. So the number of possible variable names with length at most 5 is
[itex]\sum_{k=1}^5 53 \cdot 63^{k-1} = 848,373,173[/itex] possible variable names with length at most 5.

(c) For length 1, there are only 53 such possible variable names. For length 2, the first and last characters must be the same, so there are again only 53 such possible variable names. For length 3, there are 53 * 63 variable names, and the same number of variable names of length 4. For length 5, there are 53 * 63 * 63 possible variable names. Hence, there are
[itex]53 + 53 + 53 \cdot 63 + 53 \cdot 63 + 53 \cdot 63^2 = 217,141[/itex]
possible variable names with the desired property.

Thanks!
 
Physics news on Phys.org
tylerc1991 said:

Homework Statement



I am using a book that doesn't have any solutions in it, so I would like to be sure that I am doing the problems right before I move on. The question is below:

In a programming language, a variable name must start with a letter or the underscore character, and succeeding characters must be letters, digits, or the underscore character. Uppercase and lowercase letters are considered different.
(a) How many variable names with exactly 5 characters can be formed?
(b) How many are there with at most 5 characters?
(c) How many are there with at most 5 characters, if they must read the same forwards and backwards? (note that kayak is acceptable but Kayak is not).

The Attempt at a Solution


(a) There are 53 possible characters for the first position, and 63 possibilities for the remaining 4 positions. So there are
[itex]53 \cdot 63^4 = 834,906,933[/itex]
possible 5 character variable names.

(b) For some [itex]k[/itex], there are [itex]53 \cdot 63^{k-1}[/itex] possible variable names of length [itex]k[/itex]. So the number of possible variable names with length at most 5 is
[itex]\sum_{k=1}^5 53 \cdot 63^{k-1} = 848,373,173[/itex] possible variable names with length at most 5.

(c) For length 1, there are only 53 such possible variable names. For length 2, the first and last characters must be the same, so there are again only 53 such possible variable names. For length 3, there are 53 * 63 variable names, and the same number of variable names of length 4. For length 5, there are 53 * 63 * 63 possible variable names. Hence, there are
[itex]53 + 53 + 53 \cdot 63 + 53 \cdot 63 + 53 \cdot 63^2 = 217,141[/itex]
possible variable names with the desired property.

Thanks!

Your calculations look correct to me.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 7 ·
Replies
7
Views
4K
Replies
3
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K