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Permutations and Combinations question

2018 · Shift 1 · Q26
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Permutations and Combinations question

2018 · Shift 1 · Q26

JEE AdvancedMathematicsPermutations and CombinationsNumerical+3 / −1
The number of 5 digit numbers which are divisible by 4, with digits from the set {1, 2, 3, 4, 5} and the repetition of digits is allowed, is .................
Numerical answer
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Correct answer: 625

Step-by-step Solution:

  1. Understand the conditions: We need to form 5-digit numbers that are divisible by 4. The digits must be chosen from the set {1, 2, 3, 4, 5}, and repetition of digits is allowed.

  2. Apply the divisibility rule for 4: A number is divisible by 4 if the number formed by its last two digits (the tens and units place) is divisible by 4. Let the 5-digit number be represented as d1 d2 d3 d4 d5. The condition is that the 2-digit number d4 d5 must be divisible by 4.

  3. Find the possible pairs for the last two digits (d4 d5): We need to find all two-digit numbers formed using digits from {1, 2, 3, 4, 5} that are divisible by 4. Let's list them systematically:

    • Numbers ending in 1: None are divisible by 4.
    • Numbers ending in 2: 12, 22 (not div by 4), 32, 42 (not div by 4), 52. The valid numbers are 12, 32, 52.
    • Numbers ending in 3: None are divisible by 4.
    • Numbers ending in 4: 14 (not div by 4), 24, 34 (not div by 4), 44, 54 (not div by 4). The valid numbers are 24, 44.
    • Numbers ending in 5: None are divisible by 4.

    So, the possible pairs for the last two digits (d4 d5) are: 12, 24, 32, 44, 52. There are a total of 5 possible combinations for the last two digits.

  4. Fill the remaining three digits (d1, d2, d3): Since repetition of digits is allowed, we can choose any of the 5 digits for the first three positions.

    • Number of choices for the first digit (d1): 5 (can be 1, 2, 3, 4, or 5).
    • Number of choices for the second digit (d2): 5 (can be 1, 2, 3, 4, or 5).
    • Number of choices for the third digit (d3): 5 (can be 1, 2, 3, 4, or 5).

  5. Calculate the total number of valid 5-digit numbers: Using the fundamental principle of counting (multiplication rule), the total number of such 5-digit numbers is the product of the number of choices for each position.

    Total numbers = (choices for d1) × (choices for d2) × (choices for d3) × (choices for the pair d4 d5)

    Total numbers=5×5×5×5\text{Total numbers} = 5 \times 5 \times 5 \times 5Total numbers=5×5×5×5 Total numbers=54\text{Total numbers} = 5^4Total numbers=54 Total numbers=625\text{Total numbers} = 625Total numbers=625

    Thus, there are 625 such 5-digit numbers.

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