Answer

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Hint: ince, the digits are not repeated, in units place we can have either one of $2,$ $4,$$6$. So now hundreds and tens place is to be filled. We have $5$digits remaining, so we have to select 2 digits from $5$ 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:

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 ${}^{n}{{C}_{r}}$,

where ${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$.

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 $5$ digits.

Therefore, the number of ways in which hundreds and tens place can be filled with the remaining $5$ digits is the permutation of $5$ different digits taken $2$ at a time.

${}^{5}{{P}_{2}}=\dfrac{5!}{(5-2)!}=\dfrac{5!}{3!}$.

Number of ways of filling hundreds and tens place $=\dfrac{5\times 4\times 3!}{3!}=20$

So the number of $3$-digit even numbers with $2$ at unit place$=20$.

Similarly, the number of $3$-digit even numbers with $4$ at unit place$=20$.

Number of $3$-digit even numbers with $6$ at unit place$=20$.

For total $3$-digit even numbers$=20+20+20=60$

The $3$-digit even numbers can be made using the digits $1,$ $2,$ $3,$$4,$ $5,$ $6,$ $7$ if no digit is repeated in $60$ways.

Note: Read the question carefully. Do not make silly mistakes. Also, don’t jumble while simplifying. Solve it step by step. Don’t miss any term. Take utmost care that you are going step by step way. Your concepts regarding permutations should be cleared.

Complete step-by-step answer:

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 ${}^{n}{{C}_{r}}$,

where ${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$.

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 $5$ digits.

Therefore, the number of ways in which hundreds and tens place can be filled with the remaining $5$ digits is the permutation of $5$ different digits taken $2$ at a time.

${}^{5}{{P}_{2}}=\dfrac{5!}{(5-2)!}=\dfrac{5!}{3!}$.

Number of ways of filling hundreds and tens place $=\dfrac{5\times 4\times 3!}{3!}=20$

So the number of $3$-digit even numbers with $2$ at unit place$=20$.

Similarly, the number of $3$-digit even numbers with $4$ at unit place$=20$.

Number of $3$-digit even numbers with $6$ at unit place$=20$.

For total $3$-digit even numbers$=20+20+20=60$

The $3$-digit even numbers can be made using the digits $1,$ $2,$ $3,$$4,$ $5,$ $6,$ $7$ if no digit is repeated in $60$ways.

Note: Read the question carefully. Do not make silly mistakes. Also, don’t jumble while simplifying. Solve it step by step. Don’t miss any term. Take utmost care that you are going step by step way. Your concepts regarding permutations should be cleared.

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