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JEE Main 2019
Permutations & Combinations
Permutations and Combinations
Easy

Question

n-digit numbers are formed using only three digits 2, 5 and 7. The smallest value of n for which 900 such distinct numbers can be formed, is :

Options

Solution

Key Concepts and Formulas

  • Fundamental Principle of Counting (Multiplication Rule): If there are mm ways to do one thing and nn ways to do another, then there are m×nm \times n ways to do both. This extends to any number of independent events.
  • nn-digit numbers with repetition: If we are forming nn-digit numbers using kk distinct digits, and repetition is allowed, then the total number of distinct numbers that can be formed is knk^n.
  • Inequalities: Understanding how to set up and solve inequalities to find the minimum or maximum value that satisfies a given condition.

Step-by-Step Solution

Step 1: Identify the parameters and the goal.

We are given that we need to form nn-digit numbers using only the digits 2, 5, and 7. This means we have 3 choices for each digit. Repetition is allowed. Our goal is to find the smallest value of nn such that we can form at least 900 distinct numbers.

Step 2: Determine the number of nn-digit numbers that can be formed.

Since we have 3 choices for each of the nn digits, and repetition is allowed, the total number of nn-digit numbers that can be formed is 3n3^n. This follows directly from the Fundamental Principle of Counting. Total number of distinct numbers=3n\text{Total number of distinct numbers} = 3^n

Step 3: Set up the inequality.

We want to find the smallest nn such that the number of distinct numbers formed is greater than or equal to 900. This translates to the inequality: 3n9003^n \ge 900

Step 4: Solve the inequality by testing integer values of n.

We need to find the smallest integer nn that satisfies the inequality. We test increasing values of nn:

  • For n=1n = 1: 31=33^1 = 3. Since 3<9003 < 900, n=1n=1 is not a solution.
  • For n=2n = 2: 32=93^2 = 9. Since 9<9009 < 900, n=2n=2 is not a solution.
  • For n=3n = 3: 33=273^3 = 27. Since 27<90027 < 900, n=3n=3 is not a solution.
  • For n=4n = 4: 34=813^4 = 81. Since 81<90081 < 900, n=4n=4 is not a solution.
  • For n=5n = 5: 35=2433^5 = 243. Since 243<900243 < 900, n=5n=5 is not a solution.
  • For n=6n = 6: 36=7293^6 = 729. Since 729<900729 < 900, n=6n=6 is not a solution.
  • For n=7n = 7: 37=21873^7 = 2187. Since 21879002187 \ge 900, n=7n=7 is a solution.

Since n=6n=6 is too small, and n=7n=7 is the next smallest integer, n=7n=7 is the smallest integer that satisfies the inequality.

Common Mistakes & Tips

  • Forgetting Repetition: Failing to recognize that repetition is allowed would lead to a completely different (and incorrect) approach.
  • Incorrect Inequality: Setting up the inequality as 3n=9003^n = 900 would be incorrect since we are looking for the smallest nn for which at least 900 numbers can be formed.
  • Miscalculation: Errors in calculating the powers of 3 could lead to selecting the wrong value of nn.

Summary

The problem involves finding the smallest number of digits, nn, such that at least 900 distinct numbers can be formed using the digits 2, 5, and 7 with repetition allowed. This is solved by recognizing that the number of distinct nn-digit numbers is 3n3^n, setting up the inequality 3n9003^n \ge 900, and testing integer values of nn until the inequality is satisfied. The smallest integer value of nn that satisfies the condition is n=7n=7.

The final answer is 7\boxed{7}, which corresponds to option (B).

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