Solve the given permutation problem

  • Thread starter Thread starter chwala
  • Start date Start date
  • Tags Tags
    Permutation
Click For Summary
SUMMARY

The discussion focuses on solving a permutation problem involving the arrangement of digits with specific constraints. The first approach calculates the total arrangements as 5,712 by considering the last digit as either 0 or 5, while ensuring that 0 does not occupy the first slot. The second approach also arrives at 5,712 arrangements by initially filling slots with digits not divisible by 5 and then calculating the permutations accordingly. Both methods confirm the total number of valid permutations is 5,712.

PREREQUISITES
  • Understanding of permutations and combinations
  • Familiarity with basic number theory concepts
  • Knowledge of factorial notation and its application in counting
  • Ability to analyze constraints in mathematical problems
NEXT STEPS
  • Study advanced permutation techniques in combinatorics
  • Learn about constraints in combinatorial problems
  • Explore applications of permutations in algorithm design
  • Investigate the use of factorials in probability theory
USEFUL FOR

Mathematicians, computer scientists, and students studying combinatorics or algorithm design will benefit from this discussion, particularly those interested in solving complex permutation problems.

chwala
Gold Member
Messages
2,828
Reaction score
421
Homework Statement
How many ##5##-digit numbers are there that have ##5## different digits and are divisible by ##5##?
Relevant Equations
Permutation
My First approach on this;
The last digit should be a ##0## or a ##5##. Therefore the first ##4## slots can be filled in ##_9P_4×1## ways = ##3024##.
Secondly, we also note that ##0## cannot be on the first slot as this would imply ##4## digits instead of ##5##...further the last slot has to either be a ##0## or a ##5##. Therefore we are going to have; ##8×_8P_3×1##=##2688##. Thus we shall have ##3024+2688=5712## ways.

In my second approach;

I am thinking along the lines of the ##5## slots being filled initially by digits that are not divisible by ##5## i.e
The ##5## slots can be filled in ##9×_9P_4## ways= ##27, 216## ways.

The ##5## slots can be filled by numbers that are not divisible by ##5## in ;
##8×_8P_3×8=21,504## ways.

Therefore, numbers that are divisible by ##5## can be filled in ##27,216 - 21,504=5,712## ways. Bingo! :cool:
Any feedback is highly welcome or more insight that is.
 
Physics news on Phys.org
Alternatively:

If the number ends in ##0##, three are ##9\times 8 \times 7 \times 6## possibilities.

If the number ends in ##5## there are ##8\times 8 \times 7 \times 6## possibilities.
 
  • Like
Likes   Reactions: chwala

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
Replies
3
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
Replies
4
Views
2K