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What is the probability of getting 2 chosen ranked cards (any suit). Before the dealer, when the cards are dealt alternately starting from the player?

The result is $$ \mathbb{P}(\text{player wins}) = \frac{143067}{279650} = 0.511593062757018 $$ We can use recursion with two ranks too. We can consider the deck to consist of four $0$'s, four $1$'s (...
ploosu2's user avatar
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Number of ways arrival and departure events happen in a FIFO queue.

Answer: The number of ways this can happen is $$\binom{M+N}N-\binom{M+N}{U}$$ Proof: In the window, there are exactly $N$ departures and $M$ arrivals, so there are at most $\binom{M+N}{M}$ sequences. ...
Mike Earnest's user avatar
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Probability of choosing 4 cards whose sum is 5 from a deck of 40 cards

To get a sum of five, you need exactly one 2 and three aces. The chance for 2AAA is 4/40 times 4/39 times 3/38 times 2/37. Since the 2 can be in any of four positions you multiply by 4. Now something ...
gnasher729's user avatar
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8 votes
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Probability of choosing 4 cards whose sum is 5 from a deck of 40 cards

As other responses have indicated, the first two choices are wrong. The third choice, which (like the other two choices) is attempting to use a probability-oriented approach, is also wrong, ...
user2661923's user avatar
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2 votes

Probability of choosing 4 cards whose sum is 5 from a deck of 40 cards

In the light of the comment from @user2661923 In the first case, you consider 2,2,2,2 it doesn't work. In case 2, also only one particular possibility additionally was considered not all needed as ...
Teodoras Paura's user avatar
0 votes

How many different 8 characters passwords with 2 upper-case 2 digits 4 lower-case

"...Is this correct..." Your posted analysis seems very bizarre to me. Based on the posted question: There are $~420~$ different ways that the two upper case characters, two digits, and ...
user2661923's user avatar
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2 votes
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Error in Thinking about Combinations of 9 Digits

It is true that your 2nd approach, as it is stated in the posted question, is erroneous. It is also true that the error is as stated in your answer: you are over-counting. However, your 2nd approach ...
user2661923's user avatar
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2 votes

Error in Thinking about Combinations of 9 Digits

I figured out where I went wrong while typing this out. Consider two of the strings from the erroneous answer: 8xxxxxxxx and xxxxxxxx8. If we pick 00000008 = xxxxxxxx for the first string and ...
RD Healthcare's user avatar
1 vote
Accepted

How many elements, in a set of binary strings, has exactly $k$ blocks of repeating character.

Note: the problem statement is ambiguous. I took "of any length" to mean that the blocks of $u's$ might not have the same length, and that's what the following solution computes. But that ...
lulu's user avatar
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0 votes
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How many different 8 characters passwords with 2 upper-case 2 digits 4 lower-case

With generating function: $$ [x^2y^4z^2] (26x+26y+10z)^8 \\ = [x^2y^4z^2] \sum_{a=0}^8 \sum_{b=0}^{8-a} \frac{8!}{a!b!(8-a-b)!}(26x)^a(26y)^b(10z)^{8-a-b} \\ = \frac{8!}{2!4!2!}26^2 26^4 10^2 = ...
ploosu2's user avatar
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Permutation of $n$ objects taken $r$ at a time having $p$ similar objects

Let us name the 5 red balls as R1, R2, R3, R4, R5 and the 2 black balls as B1, B2. Now, let us, temporarily, treat R1, R2, R3, R4, R5 as different balls and B1, B2 as different balls. Then, the number ...
Het Mundra's user avatar
1 vote

A mapping is selected at random from all defined on set A . If the selected mapping is injective, the probability that only one element maps on itself

If you're familiar with cycle notation and cycle structure for permutations you could approach the problem in this way: There are $5$ ways to pick an element to map to itself. The remaining four ...
paw88789's user avatar
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1 vote

A mapping is selected at random from all defined on set A . If the selected mapping is injective, the probability that only one element maps on itself

The issue is with your $3\times 3 \times 2 \times 1$, which I assume you think is the number of complete derangements of four items. A better calculation would be: The first item of four can be in ...
Henry's user avatar
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1 vote

A mapping is selected at random from all defined on set A . If the selected mapping is injective, the probability that only one element maps on itself

If only one element maps on to itself, the other $4$ mappings must be deranged. A derangement of $4$ elements, i.e. a permutation where no element is in its "proper" position (fixed), can be ...
true blue anil's user avatar
0 votes
Accepted

Binary combinations - rank and unrank

To calculate the rank of binary vector $b$, for all the places where there is $1$, add the binomial coefficients $n-1-j\choose s$ where $j$ is the index of the place and $s$ is the count of $1$'s to ...
ploosu2's user avatar
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0 votes

A game requires 2 players opposite 2 other players, with 6 people available, how many distinct games can take place?

A more pedestrian approach: you need to select 4 people out of 6, for the first one you have six possibilities, second: five etc. But then you need to rule out some permutations: one within each pair, ...
TAR86's user avatar
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0 votes

A dance class consists of 22 students, 10 women and 12 men. If 5 men and 5 women are to be chosen and then paired off, how many results are possible?

Since you are arranging pairs of men and women, you can think of them as groups. After you have 5 men and 5 women chosen, you will have 5 groups that you need to arrange. This is why you only need to ...
b-trz's user avatar
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4 votes

A game requires 2 players opposite 2 other players, with 6 people available, how many distinct games can take place?

With the tennis season heating up, we can look at it like arranging doubles tennis matches. $4$ individuals can be selected in $\binom64 = 15$ ways and the tallest among them can be paired with any ...
true blue anil's user avatar
5 votes

A game requires 2 players opposite 2 other players, with 6 people available, how many distinct games can take place?

Your pairs overlap. It's no good taking $(p_1, p_2)$ and $(p_1,p_3)$ as your two pairs, right? To repair your calculation: Having chosen one pair, there are now $4$ people left to choose from, and $\...
lulu's user avatar
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0 votes
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A question about dice combinations - cube digit pairs

You're not properly accounting for the fact that you have 12 spaces and 9 numbers, which means several numbers will appear on both dice. (Since we don't use 7, we can duplicate up to 4 numbers.) Your ...
Calvin Lin's user avatar
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6 votes

Arranging Letters in a string

Successively apply the well known "gap" and "subtraction" methods. Firstly, keep the $B's$ separate by placing them in the gaps of $-A-A-C-D-D-$ and permute the other letters, ...
true blue anil's user avatar
5 votes

Arranging Letters in a string

In your thinking there is the problem, that in the way that you do you ommit some of possible arrangements. For example in your first way you do not count in strings that have $BDB$ somewhere inside ...
Rupert Rybka's user avatar
3 votes

Arranging Letters in a string

None of your three methods will lead directly to the correct result because each time you are missing some possibilities. Your third method does not allow B and D to be neighbours. Your first method ...
Lieven's user avatar
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0 votes

Arranging pairs of friends in a row

Another way to count arrangements where at least one pair of best friends sits together is to condition on where you find the first adjacent pair of best friends as you move left to right along the ...
paw88789's user avatar
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1 vote

Arranging pairs of friends in a row

Your solution is not correct. Here's how to complete it. We first pick someone (X) to sit in seat 3 - there are $6$ choices. We now have $4$ choices for the person (Y) in seat 4. (There is no reason ...
Especially Lime's user avatar
1 vote
Accepted

Arranging pairs of friends in a row

Your answer is not correct. One way of getting a correct answer is to apply inclusion-exclusion, gluing together permutable pairs, and get At least $2$ friends together $ 2^1\binom31 5! -2^2\binom32 4!...
true blue anil's user avatar

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