There are 11 similar balls and 9 balls are numbered 1-9 and 2 balls are numbered 0. These balls are put in to a box and without replacement randomly 3 balls are taken out.
1)How many 3 digit numbers can there be ?
2)How many numbers with different digits ?
3)How many numbers with the digit '0' ?
4)If the balls taken out are put back in before choosing the next one, what are the answers for the above 3?
I am having trouble how to solve this question. Please explain how this is done.
I have tried doing a simple example like this :
I chose the digits as 1,2,0,0 and tried to create 2 digit numbers. Then there can be only 7 numbers.
00, 01, 02, 10, 12, 20, 21
But totally there can be 4P2 = 12 permutations. From that 5 are rejected because of the repeating 0.
But I can't find a way to calculate this using nCr or nPr.
As I suggested in my above comment, before ever thinking about trying to use nPr or nCr formulae you should be thinking about more fundamental concepts such as the rule of sum and the rule of product.
Reworded, the rule of sum essentially says that if we want to count how many outcomes are in a scenario, if we can break it into cases where the cases don't overlap and we count each outcome in one of the cases, then we can count the cases individually and add. In symbols, this could be phrased for finite sets as "if $X=A\cup B$ and $A\cap B=\emptyset$ then $|X|=|A|+|B|$."
Reworded, the rule of product essentially says that if we want to count how many outcomes are in a scenario, if we can uniquely describe each outcome via a sequence of steps such that at each step the number of options available does not depend on selections in previous steps (though the specific options available may indeed change), then the total number of outcomes is the product of the number of options available at each step. In symbols, this could be phrased for finite sets as "if $X=A\times B$ then $|X|=|A|\times |B|$"
For the first part of your problem, "How many 3 digit numbers can be formed" we need to first be clear what the question is asking. The phrase "3 digit numbers" is often very specifically meant to refer to the set of integers between $100$ and $999$. A number such as $012$ is considered a two digit number and is not considered a three digit number as leading zeroes do not count, though it is considered a three digit string. I will operate under the assumption that we are indeed working with counting three digit numbers where leading zeroes are not allowed, though all explanation that follows can easily be modified.
Break into cases so we can apply the rule of sum:
No zeroes are used
One zero is used
Two zeroes are used
(If balls are replaced, there is a fourth case available. Do you see what that is?)
In the first case, we count using the rule of product using the following steps:
Pick the first digit
Pick the second digit
Pick the third digit
In the first case, we have $9$ options for the first step, $8$ options for the second step, and $7$ options for the third step. Multiplying these, we get then $9\times 8\times 7$ outcomes in the first case. (The number of options decreases each time because the ball we used for that number has been removed from the bag. If the ball was replaced each time, this will of course change the result. Can you see how?)
In the second case, since we specified that exactly one zero must be used, break it down by rule of product by first deciding which location the zero occupies. Then pick the furthest left undetermined digit, and then pick the final undetermined digit. We have $2$ options for the first step (3 options if you allow leading zeroes), and then $9$ options for the second step and $8$ options for the third step. This gives $2\times 9\times 8$ outcomes in this case.
In the third case, since we specified that exactly two zeroes must be used, break it down by rule of product by first deciding which locations the zeroes occupy. Then pick the remaining digit. We have only one option for the first step (3 options if you allow leading zeroes), and then $9$ options for the second step. This gives $9$ options in this case.
This gives a combined total then of $9\times 8\times 7+2\times 9\times 8+9$ outcomes altogether.
For the second part of the question, how many of these three digit numbers we just counted don't have any repeated digits, notice that we already did most of the work for this. One of the three earlier cases will have repeated digits. (In the case that the balls are returned and reused after each pull, you'll just wind up using the same logic and methods used earlier)
The third part is also answered using our earlier work.
The fourth part will have only subtle changes to the earlier work.