From the digits 1, 2, 3, 4, 5, 6 and 7, how many 6 digit even numbers can be formed such that no digit is repeated and the second last digit (tens place) is even?

Introduction / Context:
This question is about constructing 6 digit even numbers from a given set of digits with no repetition. There is an additional constraint that the second last digit must be even. It tests understanding of positional restrictions and counting with permutations.

Given Data / Assumptions:

• Available digits: 1, 2, 3, 4, 5, 6, 7.
• We must form 6 digit numbers.
• No digit may repeat in any number.
• The number must be even, so the units digit must be even.
• The second last digit (tens place) must also be even.
• Even digits available are 2, 4 and 6.

Concept / Approach:
We work position wise. The last digit must be even and the second last digit must also be even, and these two positions must have different even digits because repetition is not allowed. Once we choose and arrange these two even digits for the last two positions, we then arrange the remaining four positions (from the remaining five digits) in any order. Finally, we multiply the counts of these independent choices.

Step-by-Step Solution:
Step 1: Identify the even digits: 2, 4 and 6.Step 2: We must fill the last two places (tens and units) with 2 distinct even digits. This is a permutation of 3 even digits taken 2 at a time.Step 3: Number of ways to choose and arrange these two even digits for the last two positions is 3P2 = 3 * 2 = 6.Step 4: After using 2 of the 3 even digits, we have 5 digits left (since we started with 7 digits total).Step 5: The first four positions (lakhs, ten thousands, thousands, hundreds) must be filled with any 4 of the remaining 5 digits, with no repetition and no additional restrictions.Step 6: The number of ways to arrange 4 positions from 5 available digits is 5P4 = 5 * 4 * 3 * 2 = 120.Step 7: Combine the choices for the last two positions and the first four positions: total numbers = 6 * 120 = 720.

Verification / Alternative check:

Why Other Options Are Wrong:

• 521 and 225: These numbers do not arise from any natural product of permutations here and likely result from miscounting positions or mixing combinations with permutations.
• 420: This undercounts and might arise if someone uses 3C2 instead of 3P2 for the last two positions, ignoring the order of the two even digits.

Common Pitfalls:
Students often mistakenly use combinations instead of permutations for positional assignments, forgetting that the order of digits matters in a number. Some also forget that digits cannot repeat and allow the same even digit in both the second last and last position. Others may incorrectly constrain the leading digit, although here there is no ban on using 1 as the first digit. Being clear on each positional rule helps avoid these errors.

Final Answer:
The number of 6 digit even numbers that can be formed under the given conditions is 720.