Fundamental Principle of counting

[Saumyojit]

How many numbers lying between 9 to 1000 can be formed with the digits 0,1,2,3,7,8 if the repetition of the digits is allowed?

For 2digit nos we have by fpc=5 *6=30 two digit nos
For 3digit nos we have by fpc=5 *6*6=180 three digit nos. Hold on

Now i have learnt from the book It was written in the introduction of permutation and combination that they are two things used in counting :
"If there is a choice=either this event or the other =we use + operator

If there is No Choice=all events must happen simultaneously (event 1 and event 2 and..event n)=we multiply all the events
For each of the event 'A' to occur in 'm' ways i.e for each of the 'm' ways ,event b can occur in 'n' ways; total no of ways=m*n ways "

Then they gave a eg--> we have 6 blue markers and 5 black markers.

CASE 1:If we choose 3 blue markers or 2 black markers then I have a choice of either 3 blue or 2blck , then i have to use + operation .


6c3+5c2=20ways of selecting 3 blue markers or i.e (+)10ways of selecting 2black markers= 30 combination i get
CASE 2(NO CHOICE):If we choose 3 blue markers and 2 black markers then I have no choice to select any one event other than to select both 3 blue and 2blck simultanelosuly, we must use * operation .

6c3*5c2=20ways of selecting 3 blue markers and i.e (*)10ways of selecting 2black markers=20*10=200 combination i get
This eg from the book i understood . Now combing back to where we left-->

For 2digit nos we have by fpc=5 *6=30 two digit nos
For 3digit nos we have by fpc=5 *6*6=180 three digit nos
Q: Now we have to include both 2digit and 3 digit nos so why we will not multiply result of two events i.e 180*30 but actually we are adding 180+30=210 to get the answer.
That means when we use " and" in this particular statment-->Now we have to include both 2digit and 3 digit nos ;so here "and" means we have to sum right? just simple meaning of and in that sense we are implying? Not that "and" of case2 where we multiply two events
@Dr.Peterson @JeffM

 

[Dr.Peterson]

Q: Now we have to include both 2digit and 3 digit nos so why we will not multiply result of two events i.e 180*30 but actually we are adding 180+30=210 to get the answer.
That means when we use " and" in this particular statment-->Now we have to include both 2digit and 3 digit nos ;so here "and" means we have to sum right? just simple meaning of and in that sense we are implying? Not that "and" of case2 where we multiply two events
@Dr.Peterson @JeffM

Do you really think that you can have a number that is both 2-digit AND 3-digit? No, you have a CHOICE: 2 digits OR 3 digits, for any given number.

I have seen this issue commonly among students: the words "and" and "or" mean different things depending on whether they apply to individual items or to sets. Here, each number is either a 2-digit number OR a 3-digit number, so you ADD. It is a union.

You can express the same idea of a union by saying you want to include both the set of 2-digit numbers AND the set of 3-digit numbers; but that "and" is not the one that means addition! It actually means there is a choice.

I think that is more or less what you are saying at the end. And as you can see, the way it has been explained to you is a little too simplified. (It also omits important ideas like mutually exclusive sets.) I never trust teaching that asks you to lean too heavily on words; it's the meaning (union and intersection) that is important here.

 

[pka]

How many numbers lying between 9 to 1000 can be formed with the digits 0,1,2,3,7,8 if the repetition of the digits is allowed?

To Saumyojit, I am truly reluctant to join this discussion. I have taught courses in counting theory for at least thirty years. Yet I find your confusion a puzzlement. Solving counting problems does not involve memorizing what formula to use on which question. Oh no no, not at all; but that is what I see in what you have posted. The way to solve these one learns to count models.
Suppose that $D=\{0,1,2,3,7,8\}$ is the set of digits used in this count.
We are to count numbers between $9~\&~1000$ that means if $N$ is counted then $N$ made of the digits in $D$ and $9<N<1000$.
Therefore, we count numbers $AX,~AXX,~\&~AXXX$ where $A\ge 1~\&~X\in D$.
Can you count each of those three types? Then why add?
One learns to break a question down to disjoint events that can be counted.