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Question

How many four-digit numbers divisible by 10 can be formed using 1, 5, 0, 6, 7 without repetition of digits?

The correct answer is

24

Counting Four-Digit Numbers Divisible by 10 Without Repetition

We are asked to find the number of four-digit numbers that can be formed using the digits 1, 5, 0, 6, and 7 without repeating any digit. Additionally, the numbers must be divisible by 10.

Understanding the Constraints for Divisibility by 10

A key constraint is that the four-digit number must be divisible by 10. A number is divisible by 10 if and only if its units digit is 0.

The available digits are 1, 5, 0, 6, and 7. Since 0 is available in this set, we can form numbers divisible by 10 by placing 0 in the units position.

Applying the Non-Repetition Rule

The digits used to form the four-digit number must not be repeated. This means once a digit is used in one position, it cannot be used in any other position.

Step-by-Step Formation of the Four-Digit Number

Let the four-digit number be represented by four positions: Thousands, Hundreds, Tens, and Units.

_ _ _ _ (Thousands, Hundreds, Tens, Units)

  1. Units Position: For the number to be divisible by 10, the units digit must be 0. There is only one choice for this position: 0.
    • Choices for Units place = 1 (the digit 0)
    • Digits remaining for other places: 1, 5, 6, 7 (4 digits)
  2. Thousands Position: The number must be a four-digit number, which means the thousands digit cannot be 0. The available digits now are 1, 5, 6, and 7 (since 0 is used in the units place). Any of these 4 digits can be placed in the thousands position.
    • Choices for Thousands place = 4 (digits 1, 5, 6, or 7)
    • Digits remaining for the hundreds and tens places = 3 digits
  3. Hundreds Position: Two digits have now been used (0 in units, and one from {1, 5, 6, 7} in thousands). There are 3 digits remaining from the original set {1, 5, 0, 6, 7}. These 3 digits can be placed in the hundreds position.
    • Choices for Hundreds place = 3
    • Digits remaining for the tens place = 2 digits
  4. Tens Position: Three digits have been used. There are 2 digits remaining from the original set {1, 5, 0, 6, 7}. These 2 digits can be placed in the tens position.
    • Choices for Tens place = 2

Calculating the Total Number of Four-Digit Numbers

To find the total number of such four-digit numbers, we multiply the number of choices for each position, following the fundamental principle of counting.

Total number of four-digit numbers = (Choices for Thousands) $\times$ (Choices for Hundreds) $\times$ (Choices for Tens) $\times$ (Choices for Units)

\text{Total numbers} = 4 \times 3 \times 2 \times 1

\text{Total numbers} = 24

Therefore, 24 distinct four-digit numbers divisible by 10 can be formed using the digits 1, 5, 0, 6, 7 without repetition.

Revision Table: Understanding Number Formation

This table summarizes the choices for each digit position when forming the four-digit number divisible by 10 without repetition.

Position Constraint Available Digits (from 1, 5, 0, 6, 7) Number of Choices
Units Must be 0 for divisibility by 10 {0} 1
Thousands Cannot be 0 (used in Units), no repetition {1, 5, 6, 7} (remaining after fixing Units) 4
Hundreds No repetition (2 digits used already) 3 digits remaining 3
Tens No repetition (3 digits used already) 2 digits remaining 2

Additional Information: Permutations and Counting Principle

This problem is an application of permutations and the fundamental principle of counting.

  • Fundamental Principle of Counting: If an event can occur in \(m\) ways and another independent event can occur in \(n\) ways, then the two events can occur in \(m \times n\) ways. We applied this principle by multiplying the number of choices for each position.
  • Permutations: A permutation is an arrangement of objects in a specific order. When forming numbers with distinct digits without repetition, the order of the digits matters, so we are dealing with permutations. The number of ways to arrange \(r\) objects from a set of \(n\) distinct objects is given by the permutation formula \(P(n, r) = \frac{n!}{(n-r)!}\). However, in this problem, we had additional constraints (units digit must be 0, thousands digit cannot be 0), so we solved it by considering the choices for each position sequentially rather than directly using the standard permutation formula on the entire set of digits for all four positions at once.
  • Divisibility Rules: Knowing basic divisibility rules is crucial for solving number formation problems. The rule for divisibility by 10 (units digit is 0) is a simple but powerful one used here. Other rules, like divisibility by 2 (units digit is even), 5 (units digit is 0 or 5), 3 (sum of digits is divisible by 3), 4 (last two digits form a number divisible by 4), etc., are also important in combinatorics problems involving number formation.
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Important Questions from Permutations and Combinations

  1. What is the number of 6-digit numbers that can be formed only by using 0, 1, 2, 3, 4 and 5 (each once); and divisible by 6 ? 

  2. Consider the following statements for a fixed natural number n:

    1. C(n, r) is greatest if n = 2r

    2. C(n, r) is greatest if n = 2r - 1 and n = 2r + 1 

    Which of the statements given above is/are correct ?

  3. Let x be the number of permutations of the word ‘PERMUTATIONS’ and y be the number of permutations of the word ‘COMBINATIONS’. Which one of the following is correct ?

  4. What is the number of ways in which 3 holiday travel tickets are to be given to 10 employees of an organization, if each employee is eligible for any one or more of the tickets?

  5. A polygon has 44 diagonals then the number of its sides is

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