Get Expert Help with the Fundamental Counting Principle

A four digit no. Is to be formed by numbers(0,2,4,6,8).Find number of such numbers which are divisible by 2 or 5 when repetition is allowed ?

License plates are made using 3 letters followed by 2 digits. How many plates can be made if repetition of letters and digits is allowed?

There are 7 red and 5 yellow fish in an aquarium. Three fish are randomly caught in a net. Find the probability that the fish were:
a) All red
b) Not all of the same color.

I have solved this question using the method for dependent events as follows:
a) $\frac{7}{12}\cdot <!--\; \cdot \; -->\frac{6}{11}\cdot <!--\; \cdot \; -->\frac{5}{10}=\frac{210}{1320}$
b) $1-<!--\; -\; -->(\frac{7}{12}\cdot <!--\; \cdot \; -->\frac{6}{11}\cdot <!--\; \cdot \; -->\frac{5}{10}+\frac{5}{12}\cdot <!--\; \cdot \; -->\frac{4}{11}\cdot <!--\; \cdot \; -->\frac{3}{10})=\frac{1170}{1320}$
Is it possible to use combinations to solve this question? If yes, how?

We throw a die a number of times. We let $Y_k$ be the number of different faces that came up after $k$ throws. Now we want $P[Y_k\le <!--\; \le \; -->5]$.

According to the answer sheet, we use the principle of inclusion-exclusion here to arrive at the following answer:
$P(Y_k\le <!--\; \le \; -->5)=(\genfrac{}{}{0ex}{}{6}{5})(\frac{5}{6}{)}^k-<!--\; -\; -->(\genfrac{}{}{0ex}{}{6}{4})(\frac{4}{6}{)}^k+(\genfrac{}{}{0ex}{}{6}{3})(\frac{3}{6}{)}^k-<!--\; -\; -->(\genfrac{}{}{0ex}{}{6}{2})(\frac{2}{6}{)}^k+(\genfrac{}{}{0ex}{}{6}{1})(\frac{1}{6}{)}^k-<!--\; -\; -->(\genfrac{}{}{0ex}{}{6}{0})(\frac{0}{6}{)}^k$ (which I believe to be equal to) $P(Y_k\le <!--\; \le \; -->5)=P(Y_k=5)-<!--\; -\; -->P(Y_k=4)+P(Y_k=3)-<!--\; -\; -->P(Y_k=2)+P(Y_k=1)-<!--\; -\; -->P(Y_k=0)$

But I just can't wrap my head around how the principle of inclusion-exclusion was exactly used here. My first thought was that it has something to do with the fact that if we see 5 faces, we've also seen at least 4 faces, etc. so that $[Y_k\le <!--\; \le \; -->4]$ is a subset of $[Y_k\le <!--\; \le \; -->5]$. I've tried to manually calculate (using the principle of inclusion-exclusion) $P(Y_k=1\cup <!--\; \cup \; -->Y_k=2\cup <!--\; \cup \; -->Y_k=3\cup <!--\; \cup \; -->Y_k=4\cup <!--\; \cup \; -->Y_k=5)$ but I don't arrive at the same answer. So far I haven't come to a good intuitive understanding yet, it would be great if someone could help me grasp this.

I've been studying stats, and currently taking my first ever engineering based stats course in college. It covers Probability extensively and other stats topics. Currently, I'm stuck on recognizing key points in a problem involving permutations / combinations vs. fundamental counting principle. I have 2 example problems and what would help the most is key things to look to recognize using the counting principle vs permutations / combinations formulas. Here's one that uses the permutations / combinations according to my student solutions manual

A friend of mine is giving a dinner party. His current wine supply includes 8 bottles of zinfandel, 10 of merlot, and 12 of cabernet (he only drinks red wine), all from different wineries.

If he wants to serve 3 bottles of zinfandel and serving order is important, how many ways are there to do this?

If 6 bottles of wine are to be randomly selected from the 30 for serving, how many ways are there to do this?

If 6 bottles are randomly selected, how many ways are there to obtain two bottles of each variety?

If 6 bottles are randomly selected, what is the probability that this results in two bottles of each variety being chosen?

If 6 bottles are randomly selected, what is the probability that all of them are the same variety?

Here's one that uses counting principle

The composer Beethoven wrote 9 symphonies, 5 piano concertos (music for piano and orchestra), and 32 piano sonatas (music for solo piano).

a) How many ways are there to play first a Beethoven symphony and then a Beethoven piano concerto?

b) The manager of a radio station decides that on each successive evening (7 days per week), a Beethoven symphony will be played followed by a Beethoven piano concerto followed by a Beethoven piano sonata. For how many years could this policy be continued before exactly the same program would have to be repeated?

Any ideas would be helpful to recognize the clues in the problems.

How many natural odd numbers are between 100 and 999 that have all different digits?

There are two conditions in this question: (1)the number must be odd and (2)must have all different digits. I thought that condition 1 is more important. So, because tehere is five odd numbers, $\{1,3,5,7,9\}$, the units position has 5 choices. The tens position has 9 choices, since it must be different from the units. The hundreds position has 7 choices, because can't be 0 and must be different from the previous positions.
One gets $7\cdot <!--\; \cdot \; -->9\cdot <!--\; \cdot \; -->5=315$. But using Excel I found that there are 320 odd numbers with all different digits between 100 and 999. Where I'm wrong? Thanks

A group of pre-school children is drawing pictures ( one child is making one picture ) using 12-colours pencil set. Given that
(i) each pupil employed 5 or more different colours to make his drawing; (ii) there was no identical combination of colours in the different drawings; (iii) the same colour appeared in no more than 20 drawings,
find the maximum number of children who have taken part in this drawing activity.
( As each child can be identified with his/her unique combination of colours, the number of children can not exceed
C(5,12) + C(6,12) + C(7,12) +...+ C(12,12)
But how to NARROW it using the condition (iii) ? )

Suppose we perform 2 experiments and this first experiment can result in any one of $m$ possible outcomes. Suppose the first experiment results in outcome $i$. Then, the second experiment can result in any of $n_i$ possible outcomes $i=1,..,m$. $\mathbf{Q}\mathbf{s}\mathbf{:}$ what is the number of possible outcomes of the two experiments?

Attempt:
If we have $1,...,m$ possible outcomes and say this experiment results in one of them say $i$. and so second experiment can result in $n_i$ for each $i$ so by multiplication principle we have $n_1\cdot <!--\; \cdot \; -->n_2\cdot <!--\; \cdot \; -->..\cdot <!--\; \cdot \; -->n_m$ possible outcomes. Is this correct?

50 students traveled to Europe, last year. Of these, 12 visited Amsterdam, 13 went to Berlin, and 15 were in Copenhagen. Some visited two cities: 3 visited both Amsterdam and Berlin, 6 visited Amsterdam and Copenhagen, and 5 visited Berlin and Copenhagen. But only 2 visited all three cities.
Question : How many students visited Copenhagen, but neither Amsterdam nor Berlin?
I have done the graph and I think the answer is 6 but I would like to learn how to compute it.

In my textbook MP is strictly reserved to counting lists. Does what I do below to count subsets work?
Consider 3 men(Ace, Bob, Corry) and 3 women(Ann, Beth, Candace). Suppose we need to choose a team with 2 men and 2 women in it. An example team is $\{\text{Ace, Ann, Beth, Bob}\}$ which is the union of $\{\text{Ace, Bob}\}\{\text{Beth, Ann}\}.$ So we simply count 2−lists whose first element is a set of two men and whose second element is a set of two women. We get the same number of 2−lists if their first element is a set of two women. There are $(\genfrac{}{}{0ex}{}{3}{2})=8$ two-men subsets and that many two-women subsets. So there are 8 choices for the first element and 8 choices for the second one. In all there are $8^2={(\genfrac{}{}{0ex}{}{3}{2})}^2=64$ two element lists = teams with two men and two women in each.

I believe this question involves the rule of sum and the fundamental counting principle. I think my logic here is wrong, but I hope you could correct it.
(26 P 2 + 26 P 3)(10 P 4) = 8.19*10^7

How many four-digit numbers can be formed from the set $\{0,1,2,3,\dots <!--\; \dots \; -->,10\}$ if zero cannot be the first digit and the given conditions are to be satisfied

1. Repetitions are allowed and the number must be even.
2. Repetitions are allowed and the number must be divisible by 5.
3. The number must be odd and less than 4000 with repetition allowed.

a. my solution is 2*10*10*10= 2000 because 2 is a even number and there are 10 numbers excluding 0 in set {0,1,2,3..10} and it is 4 digits that's why 2*10*10*10 b.same in letter A but 2 is changed into 5 because it must be divisible by 5 so it is 5*10*10*10=5000 c.same to A and B... Only I changed it to 3 so my solution is 3*10*10*10 = 3000

I'm trying to apply the counting principle to the following:
"Of 300 people: 35 - bicycle and car. 40 - car and bus. 60 - bicycle and bus. 90 - bicycle. 70 - car. 105 - bus. 25 - bicycle, car, and bus."
I just don't know how this adds up to 300. If i apply the principle, I get:
90 (bicycle) + 105 (bus) + 70 (car) - 35 (bicycle and car) - 40 (car and bus) - 60 (bicycle and bus) + 25 (bicycle, car, and bus).
This equals to 155.
I don't understand how I can get 300 participants from the above, can anyone please help?

A subway car has 10 individual seats, with 5 in front and 5 in back. From 10 ​passengers, 4 prefers front seat, 3 prefers back seat and the others have no preference. In how many ways can passengers be seated, respecting preferences?
Attempt: I want to solve using only the notion of Fundamental principle of counting:

Let $F$ be each of the 4 passengers who want to sit in the front, $N$ the passengers who have no preference, and $C$ the passengers who want to sit back. Now, since we have 32˙ possibilities to take a passenger without preference (it will be multiplied by the final result) we will take one of the passengers N to sit in the front seat, we will have, by the multiplicative principle:
$5\cdot <!--\; \cdot \; -->4\cdot <!--\; \cdot \; -->3\cdot <!--\; \cdot \; -->2\cdot <!--\; \cdot \; -->1$
Now $N-<!--\; -\; -->1=2$ passengers without preference are left, and they will be exchanged with passengers who want to sit back:
$5\cdot <!--\; \cdot \; -->4\cdot <!--\; \cdot \; -->3\cdot <!--\; \cdot \; -->2\cdot <!--\; \cdot \; -->1$
But if you go on like that, and at the end of it all, multiplying it by 6 I don't get the right result. What am I wrong?
The answer is 43200

In a game, all points in the (x, y) plane with coordinates obeying $x,y\in <!--\; \in \; -->\mathbb{Z}$ are labelled as belonging to one of three players, i.e., either to Alice, Bob or Carol.
Show that one of the players will possess four points whose vertices form a rectangle.
Here is the problem I've been thinking for days. It seems easy since the coordinates are unlimited. But it is also hard to find pigeonholes I want to divide.
I've encountered several problems like finding a parallelogram on a $n\times <!--\; \times \; -->n$ chessboard with $2n$ pawns. It is relatively simpler to me.

Three friends agreed to meet at 8 P.M. in on of the Spanish restaurants in town. Unfortunately they forgot to specify the name of the restaurant. If there are 10 Spanish restaurants
(a) find the number of ways they could miss each other
(b) find the number of ways they could meet
(c) find the number of ways at least two of them meet each other.

I got 10 * 3 = 30 ways for (a) for (b), I got 10 ways. how do I do (c)?

We are supposed to use the Inclusion-Exclusion principle to solve: "how many bridge hands contain exactly 3 clubs, or exactly 5 diamonds or exactly 3 aces?" I know you do something like: |3 clubs|+|3aces|+|5dimonds| - |3clubs AND 3 aces| - |3clubs AND 5 diamonds| -...... +|all 3 intersected| But I'm really struggling with find the cases where the aces may or may not be a club/diamond.

I'm sorry for the basic question, but I can't figure this out.
Suppose you want to find how many even, non-repeating 3-digit numbers there are. If you choose the last digit to be non-zero, you have $(8)(8)(5)=320$ possibilities. If you choose the last digit to be zero, you have $(9)(8)(5)=360$ possibilities. What should I do?

I have this question,
The number of different words that can be formed using all the letters of the word “MATHEMATICS” are?
I solved using the logic that if there are total 11 places so at all the 11 places all the 11 words can come, which gives me 11 to the power 11which is then divided by $2!\ast <!--\; \ast \; -->2!\ast <!--\; \ast \; -->2!$\ due to the words M, T and A appearing 3 times.
But the actual solution is $11!/(2!\ast <!--\; \ast \; -->2!\ast <!--\; \ast \; -->2!)$, why did they consider “not repetition” ?

Note : This is not homework, it is self-study.

An employer interviews eight people for four openings in the company. Three of the eight people are women. If all eight are qualified, in how many ways could the employer fill the four positions if a)the selection is random and b)exactly two are women?

Part a) is already taken care of, now for part b). My reasoning goes like this:
There must be 2 women chosen, so imagine that for the first position we choose a woman. There's 3 ways to choose. For the second position we choose another woman, there's 2 ways to choose. For the third position we choose a man, there's 5 ways to choose. For the fourth position we choose another man, there's 4 ways to choose. Then we multiply: $3\cdot <!--\; \cdot \; -->2\cdot <!--\; \cdot \; -->5\cdot <!--\; \cdot \; -->4=120$. But the answer in the book is 30.