Math Problem Statement
How many 9-digit numbers can be created using the digits from the number 214432213, such that the even places only accommodate the even digits in the resultant number? [units place is an odd place, tens place is an even place and so on]
Solution
To solve this problem, let's analyze the constraints step-by-step.
Given Data and Observations
-
Available Digits: The number provided is 214432213.
- We have the digits: .
- Frequency of each digit:
- 2 appears 3 times.
- 1 appears 2 times.
- 4 appears 2 times.
- 3 appears 2 times.
-
Objective: Construct a 9-digit number such that:
- Only even digits occupy the even places (2nd, 4th, 6th, and 8th).
- The odd places (1st, 3rd, 5th, 7th, and 9th) can be occupied by either even or odd digits.
-
Identifying Even and Odd Digits:
- Even digits are 2 and 4.
- Odd digits are 1 and 3.
-
Even Places Requirement:
- Since there are four even places (2nd, 4th, 6th, and 8th), we need to fill these with the even digits.
- We have a total of 5 even digits (three 2's and two 4's), which is sufficient to fill the four even places.
Step 1: Choose Digits for Even Places
Since there are 5 available even digits and 4 even positions, we need to:
- Select 4 out of these 5 even digits.
The combinations for this are: After choosing 4 specific digits, we can arrange them in the 4 even places in 4! ways.
Thus, for the even places:
Step 2: Arrange Digits in Odd Places
After filling the even places with 4 chosen even digits, we are left with 5 remaining digits. These will include:
- Any leftover even digit (from the original selection of 5 even digits).
- All odd digits (total of 4 odd digits initially).
Since we have 5 remaining digits for the 5 odd places (1st, 3rd, 5th, 7th, 9th), we can arrange them in 5! ways.
Thus, for the odd places:
Step 3: Total Number of Valid 9-Digit Numbers
Now, we can find the total number of valid 9-digit numbers by multiplying the number of ways to arrange digits in the even places and the odd places:
Conclusion
The total number of 9-digit numbers that can be created under the given constraints is:
Would you like further clarification on any part of the solution?
Here are 5 additional questions to explore:
- What if the number had a different distribution of even and odd digits?
- How would the solution change if there were only 3 even places?
- What if we required specific digits in certain positions?
- How does this type of combinatorial problem generalize to numbers of different lengths?
- How would the solution be affected if duplicates of each digit were not allowed?
Tip: In combinatorics, breaking down complex constraints into independent, smaller parts can simplify counting.
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Math Problem Analysis
Mathematical Concepts
Combinatorics
Permutation and Combination
Probability Constraints
Formulas
Combination formula: C(n, r) = n! / (r! * (n - r)!)
Permutation formula: n! for arranging r items out of n
Theorems
Fundamental Counting Principle
Suitable Grade Level
Grades 10-12
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