Permutations combinatorics problem?

Click For Summary
SUMMARY

The problem involves calculating the number of unique four-digit telephone extensions with no repeated digits, where the first digit cannot be zero. The solution correctly identifies 9 options for the first digit and applies the permutation formula P(9,3) to determine the remaining digits. The final calculation yields 4,536 different extensions, confirming the accuracy of the approach and result.

PREREQUISITES
  • Understanding of permutations, specifically P(n,r) = n!/(n-r)!
  • Basic knowledge of factorials and their calculations
  • Familiarity with combinatorial principles
  • Concept of digit restrictions in numerical problems
NEXT STEPS
  • Study advanced combinatorial problems involving restrictions
  • Learn about variations of the permutation formula for different scenarios
  • Explore applications of permutations in real-world contexts
  • Investigate combinatorial proofs and their significance in mathematics
USEFUL FOR

Students studying combinatorics, educators teaching mathematical principles, and anyone interested in solving permutation-related problems.

SMA_01
Messages
215
Reaction score
0

Homework Statement



A telephone extension has four digits, how many different extensions are there with no repeated digits, if the first digit cannot be zero?

Homework Equations



P(n,r)=n!/(n-r)!


The Attempt at a Solution



For the first digit, there are 9 possibilities (because no zero)
For the last 3 digits I used P(9,3)=9!/6!=9x8x7

So, in the end my result was: 9x9x8x7= 4,536 different extensions...

I'm just wondering if I was correct?
 
Physics news on Phys.org
SMA_01 said:

Homework Statement



A telephone extension has four digits, how many different extensions are there with no repeated digits, if the first digit cannot be zero?

Homework Equations



P(n,r)=n!/(n-r)!


The Attempt at a Solution



For the first digit, there are 9 possibilities (because no zero)
For the last 3 digits I used P(9,3)=9!/6!=9x8x7

So, in the end my result was: 9x9x8x7= 4,536 different extensions...

I'm just wondering if I was correct?

Seems fine to me.
 

Similar threads

Replies
32
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
13K
Replies
8
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K