Math Problem Statement

How many number of at least three different digit numbers can be formed from the integers 1,2,3,4,5,6?

Solution

To determine the number of numbers with at least three different digits that can be formed from the integers 1, 2, 3, 4, 5, and 6, we need to consider combinations of digits and permutations of those combinations.

Step-by-Step Solution

  1. Identify Total Digits and Constraints:

    • We have 6 digits: {1, 2, 3, 4, 5, 6}.
    • The number must have at least three different digits.
    • Digits can be repeated in different positions, but the digits themselves must be unique within each number.
  2. Calculate the Number of Valid Numbers:

    • Calculate for 3-digit, 4-digit, 5-digit, and 6-digit numbers separately and then sum the results.

Number of 3-digit numbers:

  • Choose 3 digits from 6: (63)\binom{6}{3}
  • Permute those 3 digits: 3!3!

(63)×3!=20×6=120\binom{6}{3} \times 3! = 20 \times 6 = 120

Number of 4-digit numbers:

  • Choose 4 digits from 6: (64)\binom{6}{4}
  • Permute those 4 digits: 4!4!

(64)×4!=15×24=360\binom{6}{4} \times 4! = 15 \times 24 = 360

Number of 5-digit numbers:

  • Choose 5 digits from 6: (65)\binom{6}{5}
  • Permute those 5 digits: 5!5!

(65)×5!=6×120=720\binom{6}{5} \times 5! = 6 \times 120 = 720

Number of 6-digit numbers:

  • Choose 6 digits from 6: (66)\binom{6}{6}
  • Permute those 6 digits: 6!6!

(66)×6!=1×720=720\binom{6}{6} \times 6! = 1 \times 720 = 720

Total Sum:

120(3-digit)+360(4-digit)+720(5-digit)+720(6-digit)=1920120 (3 \text{-digit}) + 360 (4 \text{-digit}) + 720 (5 \text{-digit}) + 720 (6 \text{-digit}) = 1920

Conclusion

The total number of numbers with at least three different digits that can be formed from the integers 1, 2, 3, 4, 5, and 6 is 1920\boxed{1920}.

Would you like further details or have any questions?

Here are five related questions you might find interesting:

  1. How many 4-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5?
  2. How many different ways can 5 people be seated in a row?
  3. How many different permutations of the word "MATH" are there?
  4. How many different ways can you select 2 books from a set of 10 books?
  5. How many different ways can you arrange the letters in the word "ALGEBRA"?

Tip: Remember, the factorial function (n!) is the product of all positive integers up to n, and it's crucial for calculating permutations.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Permutations
Factorials

Formulas

Permutation formula: nPr = n! / (n - r)!

Theorems

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Suitable Grade Level

Grades 10-12