how many 3-digit numbers possible from a set of : 0,3,4,5,6,7

ketanco

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from a set of : 0,3,4,5,6,7

how many three digit natural numbers can be written, where, at least two digits must be the same?

answer is 80 but i can not get there

i said 5 possibility for first digit, then take the next equal, then another 6 possibility, so 5x6 = 30, and we have 5 situations like this so it makes 5x30=150 but the answer is given as 80
 
from a set of : 0,3,4,5,6,7

how many three digit natural numbers can be written, where, at least two digits must be the same?

answer is 80 but i can not get there

i said 5 possibility for first digit, then take the next equal, then another 6 possibility, so 5x6 = 30, and we have 5 situations like this so it makes 5x30=150 but the answer is given as 80

When a problem involves the phrase "at least one", it is often helpful to first calculate the number of ways with "none". That is, how many numbers can be written in which none of the digits are the same?

Can you see how to use that?

Your work did not consider when the first and last digits are the same; and you seem to have done the same multiplication twice. (Are the 5 situations the same as the 5 possibilities?)
 
from a set of : 0,3,4,5,6,7
how many three digit natural numbers can be written, where, at least two digits must be the same?
answer is 80 but i can not get there
i said 5 possibility for first digit, then take the next equal, then another 6 possibility, so 5x6 = 30, and we have 5 situations like this so it makes 5x30=150 but the answer is given as 80
First & most importantly, a three digit numbers must not have zero in the hundredth position.
The permutation of six taken three \(\displaystyle _6\mathscr{P}_3=120\). That is the number with no digit repeated.
That is the number of three digits strings. BUT of those 20 begin with a zero. They must be removed.

There are \(\displaystyle 5\cdot 6\cdot 6=180\) possible three digit numbers. So how many have at least one digit repeated?
 
First & most importantly, a three digit numbers must not have zero in the hundredth position.
The permutation of six taken three \(\displaystyle _6\mathscr{P}_3=120\). That is the number with no digit repeated.
That is the number of three digits strings. BUT of those 20 begin with a zero. They must be removed.

There are \(\displaystyle 5\cdot 6\cdot 6=180\) possible three digit numbers. So how many have at least one digit repeated?

Yes how many ? That is my questoon. i can get 180 too.
 
Let's do it your way correctly.

Possibility type 1: All three digits are the same. But the first digit cannot be zero. So there are exactly 5 ways that can happen.

Possibility type 2: First and second digits are the same. But the first digit cannot be zero. So there are 5 ways for those two digits to be the same without being zero. And that leaves 5 remaining ways to pick the third digit so it is different. 5 * 5 = 25.

Possibility type 3: First and third digits are the same. But the first digit cannot be zero. So there are 5 ways for those two digits to be the same without being zero. And that leaves 5 remaining ways to pick the second digit so it is different. 5 * 5 = 25.

Possibility type 4: Second and third digits are the same. But the first digit cannot be zero. So there are 5 ways to pick the first digit. And that leaves 5 remaining ways to pick the second and third digit so they are the same but differ from the first digit. 5 * 5 = 25.

Now add them up.

5 + 25 + 25 + 25 = 80.
 
The other answers give you ways to find the correct solution more quickly and with less chance of error.

How many ways can you construct a 3 digit string from the 6 provided if 0 is not in the first position.

5 * 6 * 6 = 180.

From those 6 digits, how many ways can you construct a string of 3 of them such that the first digit is not zero and no digit is repeated.

5 * 5 * 4 = 100.

Therefore the number of strings without an initial zero and one or more repetitions is

180 - 100 = 80.
 
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