How many 7 digit numbers can be formed using all the digits 1, 2, 0, 2, 4, 2 and 4 exactly as given, so that no extra repetition is allowed and leading zero is not permitted?

Introduction / Context:
This question tests your understanding of permutations of a multiset (repeated digits) with an additional restriction that a valid 7 digit number cannot begin with zero. We must count all distinct permutations of the given digits and then subtract those arrangements that start with zero. This type of problem is very common in aptitude tests involving digit arrangements.

Given Data / Assumptions:
Available digits: 1, 2, 0, 2, 4, 2, 4. Total digits = 7. Digit 2 appears three times. Digit 4 appears twice. Digits 0 and 1 appear once each. We must form valid 7 digit numbers, so the leading digit cannot be 0.

Concept / Approach:
First, we count the total number of distinct permutations of all 7 digits treating repeated digits properly. This uses the multiset permutation formula 7! divided by factorials of the counts of repeated digits. Then, we separately count how many of these permutations start with 0 by fixing zero in the first position and permuting the remaining 6 digits. Subtracting the bad arrangements (with leading zero) from the total gives the count of valid 7 digit numbers.

Step-by-Step Solution:
Step 1: Count total permutations of the 7 digits without any restriction. Step 2: Total permutations = 7! / (3! * 2!) because digit 2 repeats three times and digit 4 repeats twice. Step 3: 7! = 5040, 3! = 6 and 2! = 2, so denominator = 6 * 2 = 12. Step 4: Total permutations = 5040 / 12 = 420. Step 5: Now count permutations that start with 0. Fix 0 in the first position. Step 6: Remaining digits are 1, 2, 2, 2, 4, 4 (six digits with 2 repeated three times and 4 repeated twice). Step 7: Permutations of remaining digits = 6! / (3! * 2!). Step 8: 6! = 720 and 3! * 2! = 6 * 2 = 12, so we get 720 / 12 = 60. Step 9: These 60 arrangements are invalid 7 digit numbers because they begin with 0. Step 10: Valid 7 digit numbers = total permutations - invalid leading zero permutations = 420 - 60 = 360.

Verification / Alternative check:
As a quick check, note that the total count 420 includes all possible ways to order the 7 digits, regardless of the first digit. Since 1, 2 and 4 are non zero, there should be a substantial number of valid numbers. Excluding 60 sequences that start with 0 leaves 360, which is less than the total but still a large fraction, which is reasonable. A more detailed check could group valid numbers by the first digit (1, 2 or 4) and verify symmetry ideas, but the computed arithmetic from factorial formulas is already very reliable.

Why Other Options Are Wrong:
The value 120 is far too small and might come from misusing factorials or ignoring repeated digits. The value 240 could arise from subtracting a wrong number of invalid arrangements. The value 424 is larger than the maximum possible distinct permutations of 420, so it cannot be correct. Only 360 correctly equals 420 minus 60 and respects both the multiset permutations and the leading digit restriction.

Common Pitfalls:
Common mistakes include ignoring that digit 2 and digit 4 are repeated, leading to using 7! instead of 7! / (3! * 2!). Another frequent error is forgetting to remove arrangements that start with 0, which would allow invalid 7 digit numbers with a leading zero. Some students also incorrectly divide by 2 a second time or mix up the counts of repeated digits when calculating 6! / (3! * 2!). Carefully identifying repeated digits and applying constraints step by step prevents these issues.

Final Answer:
The number of distinct 7 digit numbers that can be formed is 360.