# Tag Info

### 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 (...
• 9,523

### 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. ...
• 77.4k

### 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 ...
• 10.2k
Accepted

### 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, ...
• 36.8k

### 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 ...

### 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 ...
• 36.8k
Accepted

### 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 ...
• 36.8k

### 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 ...
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 ...
• 71.5k
Accepted

### 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 = ...
• 9,523

### 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 ...
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 ...
• 40.6k
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 ...
• 158k
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 ...
• 42.6k
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 ...
• 9,523

### 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, ...
• 103

### 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 ...
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 ...