How many 4-digit even numbers can be formed from the digits 5 6 7 8 9 if no digits are repeated


The $7$ digits $1, 2, 3, 5, 6, 8, 9$ are to be used to make $5$ digit numbers with each being used not more than once in a number. How many numbers can be made which are more than 60 000 AND even? I got something like $3\cdot 3\cdot 5P_3 = 540$. But I think it's wrong because there are numbers that we can choose which are both even and $\geq 6$.


Without considering different cases.

This is another way to get the same answer as already answerd above.

Here we start by finding the total amount of the four-digit numbers with distinct digits, then finding the amount of odd digits, filling the same criteria, and last, we subtract the odd numbers from the total, to find the even numbers filling the criteria.

We have totally seven digits. We know that the digit zero cannot be placed at the position representing the thousand position. That leaves us with six digits to choose from for this position and that can be made in

$P(6,1) = \frac{6!}{(6-1)!} = \frac{6!}{5!}=6$, different ways.

Now, we have three positions left to fill and six digits to choose from, including the digit zero, which can be placed anywhere in the remaining positions. This choice can be done in

$P(6,3) = \frac{6!}{(6-3)!} = \frac{6!}{3!}=6 \cdot5 \cdot4$, different ways.

Finally, with distinct digits, there is

$6 \cdot6 \cdot5 \cdot4 =720$

four-digit numbers to be constructed.

We know that the amount of odd numbers plus the amount of even numbers equal the total amount of the 720 four-digit numbers.

The odd numbers are 1, 3 and 5. The question to be asked is how many of the 720 are odd?

The digit to fill the unit position can only be chosen from the digits 1, 3 or 5, and this choice can be made in

$P(3,1)=\frac{3!}{(3-1)!}=\frac{3!}{2!}=3$, different ways.

To choose the digit filling the thousand position, we have five valid digits to choose from, since the zero digit is not valid for this position. The choice can be made in

$P(5,1)=\frac{5!}{(5-1)!}=\frac{5!}{4!}=5$, different ways.

Now we are left with two positions to fill, the hundred and tenth position, and five digits to choose from, now including the zero digit. The choice for these two positions can be made in

$P(5,2)=\frac{5!}{(5-2)!}=\frac{5!}{3!}=5 \cdot4=20$, different ways.

The total number of odd four-digit numbers, with distinct digits are

$5 \cdot5 \cdot4 \cdot3 =300$.

Now, we can answer the question how many of these 720, four-digit numbers, are even, by the subtraction


The even numbers are 420.

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How many 4-digit even numbers can be formed from the digits 5 6 7 8 9 if no digits are repeated

Text Solution


Answer : 12

Solution : Let the 4-digit number is ending with 6. <br> <img src="" width="80%"> <br> The remaining three places can be filled using the three-digits 5, 7, 8. <br> 5786, 5876, 7586, 7856, 8576, 8765 <br> Similarly, the number if 4- digit numbers ending with eight is 6. <br> Hence, we get tweleve, 4-digit even combinations with the given digits.