Question

How many $3$-digit even numbers can be made using the digits 1, 2, 3, 4, 5, 6, 7 if no digit is repeated?

Answer 1

Hint: Since, the digits are not repeated, in the unit's place we can have either one of 2, 4, 6. So now, hundreds and tens of places are to be filled. We have 6 digits remaining, so we have to select 2 digits from 6 different digits taking 2 in the units place for the first case. After that, you will get a number of 3-digit even numbers with 2 at unit place. Similarly you will get such numbers for 4 and 6 as well. Find the numbers and add all the three cases. You will get the answer. Try it.

Complete step-by-step answer:
Even numbers are integers that are exactly divisible by 2, whereas an odd number cannot be exactly divided by 2. The examples of even numbers are 2, 4, 6 etc.
An even number is an integer that can be divided by two and remain an integer or has no remainder.
An integer that is not an even number is an odd number. If two divides an even number, the result is another integer. On the other hand, an odd number, when divided by two, will result in a non-integer.
Since even numbers are integers, negative numbers can be even.
The formula for permutations with repeated elements is as follows when $k$ out of $n$ elements are indistinguishable. So for example, if we have $k$ copies of the same book, the number of different permutations possible is $\dfrac{n!}{k!}$.
Sometimes, we want to count all of the possible ways that a single set of objects can be selected without regard to the order in which they are selected.
A combination is a selection of all or part of a set of objects, without regard to the order in which they were selected. This means that $xyz$ it is considered the same combination $zyx$.
The number of combinations of $n$ objects taken $r$ at a time is denoted by,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$
We know that a digit is even when the number at units place is even.
At the units place we can have either 2, 4 or 6.
Since the digits cannot be repeated in the 3-digit numbers and units place is already occupied with a digit (which is even), the hundreds and tens place is to be filled by the remaining 6 digits. Therefore, the number of ways in which hundreds and tens place can be filled with the remaining 6 digits is the permutation of 6 different digits taken 2 at a time.
${}^{4}{{P}_{2}}=\dfrac{6!}{(6-2)!}=\dfrac{6!}{4!}$.
Number of ways of filling hundreds and tens place$=\dfrac{6\times 5\times 4!}{4!}=30$
So, the number of 3-digit even numbers with 2 at unit place is 30.
Similarly, the number of 3-digit even numbers with 4 at unit place is 30.
Number of 3-digit even numbers with 6 at unit place is 30.
For total 3-digit even numbers,
$=30+30+30=90$
The 3-digit even numbers can be made using the digits 1, 2, 3, 4, 5, 6, 7 if no digit is repeated in 90 ways.

Note: Read the question carefully. Do not make silly mistakes. Also, do not jumble while simplifying. Solve it systematically. Do not miss any term. Take utmost care that you are going step-by-step. Your concepts regarding permutations should be cleared.
Possible mistake made by students is, once they think the unit place should consist of 2, 4, 6, then when calculating the permutations they ignore all these 3 digits and consider only 1, 3, 5, 7 i.e., 4 digits for tens place. This will lead to wrong answers.