Permutation and combination homework

Click For Summary
SUMMARY

The discussion focuses on calculating the number of three-digit odd numbers that can be formed using the digits 1, 2, 3, 4, 5, 6, and 7 under two conditions: without repetition and with repetition. For the scenario without repetition, the correct answer is 120, derived from the formula 6P2 multiplied by 4 (the choices for the last digit). For the scenario with repetition, the answer is 196, determined by considering the choices for each digit position sequentially.

PREREQUISITES
  • Understanding of permutations and combinations
  • Familiarity with basic number theory
  • Knowledge of odd and even number properties
  • Ability to apply factorial calculations
NEXT STEPS
  • Study the principles of permutations and combinations in depth
  • Learn how to calculate factorials and their applications in combinatorial problems
  • Explore advanced counting techniques, including the pigeonhole principle
  • Practice problems involving digit arrangements and constraints
USEFUL FOR

Students studying combinatorics, educators teaching mathematics, and anyone interested in solving permutation and combination problems.

jinx007
Messages
60
Reaction score
0
Find how many 3 digits odd number that can be obtained from the digit 1,2,3,4,5,6,7 if,

1/ Repetition of digits not allowed
2/ Repetition of digits allowed

my work

1/ --- the last i digit i have control 1,3,5,7

so 2 remaining digits = 6 P 2 x 4 (4ways)

= 120

2/ i AM STUCK

In fact, i don't really know whether the method is correct,i am a bit confuse, HELPPPP

ANSWER ACCORDING TO BOOKLET :
1/ 120

2/ 196
 
Physics news on Phys.org
For the second question, how many choices for the rightmost digit do you have?
After that, how many choices for the middle digit? And the first? So...
 
nvm i gave away too much for a homework question
 

Similar threads

How many ways one can put prime numbers to form 3 digit NIP?
  • · Replies 18 ·
Replies
18
Views
3K
Graduate Mapping and Recovering Combinations: A Challenge in Combination Theory
  • · Replies 11 ·
Replies
11
Views
4K
How Many 4-Digit Permutations Greater Than 5364 Can Be Formed?
  • · Replies 1 ·
Replies
1
Views
2K
How Many 5-Digit Numbers from 4, 5, 6, 8, 9 are Divisible by 8?
  • · Replies 4 ·
Replies
4
Views
2K
Questions regarding permutations .
  • · Replies 8 ·
Replies
8
Views
3K
Probability of Non-Adjacent Zeros in 11-Digit Integer with 1 and 0 Digits
  • · Replies 59 ·
2
Replies
59
Views
5K
Solving Permutation Questions: 4-Digit Even Numbers and Letter Arrangements
  • · Replies 7 ·
Replies
7
Views
7K
A simple combinatorics problem
  • · Replies 7 ·
Replies
7
Views
2K
2 Questions about Counting & Permutations
  • · Replies 6 ·
Replies
6
Views
3K
Solving Odd Digit Divisibility by Five using Permutations
  • · Replies 8 ·
Replies
8
Views
2K