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Ivan
2 years ago
10

How many three-digit natural numbers can be formed using only the even digits, if the numbers can contain each of these digits o

nly once?
Mathematics
1 answer:
ICE Princess25 [194]2 years ago
6 0

Answer:

48

Step-by-step explanation:

Solution 1:

There are 5 even digits: 0, 2, 4, 6, and 8, and 5\cdot 4\cdot 3=60 ways to arrange any three of them in order: 5 ways to pick the first digit, 4 ways to pick the second digit, and 3 ways to pick the third.

Another way to think about this is there are 5\choose 3 ways to pick any three of the digits, and 3! ways to rearrange them, so there are {5\choose 3 }\cdot 3!=60 ways to pick and arrange 3 of the digits.

Then, we have to subtract cases where 0 is the first digit, because that would give us numbers such as 028 or 042 which are not three-digit natural numbers.

When 0 is the first digit, there are 4 possibilities for the second digit and 3 possibilities for the third digit, so there are 4\cdot3=12 ways for 0 to be the first digit.

These are the cases we don't want, so we subtract them from the total cases we can arrange any three of the digits. This gives us

60-12=\boxed{48}

total ways.

Solution 2:

We split the possible arrangements into two different cases: when we use the digits without 0 and when we do use the digit 0.

Case 1: we use 0:

We first pick the three numbers we use: one of them has to be 0, so there are {4\choose 2}=6 ways to pick the other two numbers.

Then, since 0 cannot be the first digit, there are 2 ways to pick the first digit. We used a digit for the first number, so there are 2 more numbers we can choose from, including 0, for the second digit. There is only one possibility for the last digit, so we have 2\cdot2\cdot1=4 ways to pick the order.

So, multiplying this by the number of ways there are to pick the numbers, there are 6\cdot4=24 ways for this case.

Case 2: we don't use 0:

If we don't use 0, there are 4 digits we can choose from: 2, 4, 6, and 8. There are 4 ways to choose the first digit, 3 ways to chose the second digit, and 2 ways for the third.

So, we have 4\cdot3\cdot2=24 ways for this case.

Combining the cases, there are 24+24=\boxed{48} total possible ways to form a three-digit natural number using only the even digits.

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What is the square root of m^6?
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Answer:
If m is nonnegative (ie not allowed to be negative), then the answer is m^3
If m is allowed to be negative, then the answer is either |m^3| or |m|^3

==============================

Explanation:

There are two ways to get this answer. The quickest is to simply divide the exponent 6 by 2 to get 6/2 = 3. This value of 3 is the final exponent over the base m. Why do we divide by 2? Because the square root is the same as having an exponent of 1/2 = 0.5, so
sqrt(m^6) = (m^6)^(1/2) = m^(6*1/2) = m^(6/2) = m^3
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---------------------------

A slightly longer method is to break up the square root into factors of m^2 each and then apply the rule that sqrt(x^2) = x, where x is nonnegative

sqrt(m^6) = sqrt(m^2*m^2*m^2)
sqrt(m^6) = sqrt(m^2)*sqrt(m^2)*sqrt(m^2)
sqrt(m^6) = m*m*m
sqrt(m^6) = m^3
where m is nonnegative

------------------------------

If we allow m to be negative, then the final result would be either |m^3| or |m|^3. 

The reason for the absolute value is to ensure that the expression m^3 is nonnegative. Keep in mind that m^6 is always nonnegative, so sqrt(m^6) is also always nonnegative. In order for sqrt(m^6) = m^3 to be true, the right side must be nonnegative.

Example: Let's say m = -2
m^6 = (-2)^6 = 64
sqrt(m^6) = sqrt(64) = 8
m^3 = (-2)^3 = -8
Without the absolute value, sqrt(m^6) = m^3 is false when m = -2

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