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

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

$720-300=420$.

The even numbers are 420.

Okay so we have ___ ___ ___ ___ to fill in. Since the number is even, the last blank must be an even number (0,2,4,6,or 8). So there are 5 choices for the last blank. For the first blank, we cannot have a 0 (as it would not be a true 4 digit number) and we cannot have the digit that was already placed in the 4th blank. This means we have 8 options. For the second blank, we can now have 0, but we cannot have either of the two numbers that have already been placed. This means we have 8 options. Similarly, for the third blank we cannot have any of the other 3 numbers that are chosen so we have 7 options.

Using the product rule we get that there are

Hopefully this helps, let me know if you are still unsure.

Answer:

120

Step-by-step explanation:

Let us represent the 4-digit even numbers with blanks:

_ _ _ _

The blanks also represent how many numbers we can put into that spot.

An even number is a number whose ones-digit are any of 2, 4, 6, 8 or 0. Since the problem asks us to form even numbers, we need to satisfy that condition first before we go on. In forming numbers, we need to consider first those with restrictions. After all the restrictions are satisfied, we can go on the other digits.

There are 6 numbers given to us. We are only given 1, 3, 5, 6, 8, and 9. For a number to be even, it can only end with 6 or 8. This means that the ones-digit of the number only has 2 possible digits.

_ _ _ 2

That is the only restriction we have. Now we can go to the thousands-digit and list how many digits we can put there.

Since we already used 1 digit to put on the ones-place, it leaves us 5 digits to fill the thousands-digit.

5 _ _ 2

The hundreds-digit can be filled with 4 numbers, then the tens-digit can be filled with 3 numbers.

5 4 3 2

The blanks represent how many numbers can be used to fill that spot. Via the Fundamental Principle of Counting (FPC), product rule, if there are x ways of doing something, and y ways of doing another, there are x*y ways of doing both things.

With that knowledge, we know that there are 5 ways of filling up the thousands-digit, 4 ways for the hundreds-digits, 3 ways for the tens-digit, and 2 ways for the ones-digit. We multiply them to get the number of ways to get the 4-digit number.

There are 120 ways to form 4-digit even numbers from the given digits.

For more information about FPC and forming numbers with digits, click here:

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Text Solution

1210118

Answer : 12

Solution : Let the 4-digit number is ending with 6. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PS_MATH_VI_C01_E04_013_S01.png" 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.