In how many ways can b blue balls and r red balls be distributed in n distinct boxes?
122439
I have 4 unlabeled (identical) blue balls and 2 identical redz balls and 5 different (distinct) boxes. In how many ways all balls can be placed in these 5 boxes?
62763
I was wondering if someone could critique my argument here. The problem is to find the probability where exactly 2 people in a room full of 23 people share the same birthday.  My argument is that there are 23 choose 2 ways times $\displaystyle \frac{1}{365^{2}}$ for 2 people to share the same birthday. But, we also have to consider the case involving 21 people who don't share the same birthday. This is just 365 permute 21 times $\displaystyle \frac{1}{365^{21}}$. To summarize:  $$\binom{23}{2} \frac{1}{365^2} \frac{1}{365^{21}} P\binom{365}{21}$$
39445
Given n distinct objects, there are $n!$ permutations of the objects and $n!/n$ "circular permutations" of the objects (orientation of the circle matters, but there is no starting point, so $1234$ and $2341$ are the same, but $4321$ is different).  Given $n$ objects of $k$ types (where the objects within each type are indistinguishable), $r_i$ of the $i^{th}$ type, there are   \begin{equation*} \frac{n!}{r_1!r_2!\cdots r_k!} \end{equation*}  permutations.  How many circular permutations are there of such a set?
66702
I was solving some combinatorics question and I came upon one I've been trying to solve for 2 days and I'm curios how to solve it.  It is known that in every set of 15 cars, at least 3 were manufactured from the same country (there can also be another 3 (or more) cars in the set which came from another country). prove that in every set of 100 cars, 15 were made manufactured in the same country.  Thanks to anyone who helps!
59448
The question is: n balls are distributed into n holes at random. What is the probability that exactly one hole remains empty. I came up with $$P\left(A\right)=\frac{\:\dbinom{n\:}{1}\:\dbinom{n\:-1}{1}\:\left(n-2\right)!}{n^n}$$ but was told I’m way way off. Yes, I do realize I’m awful at probability. Thanks for the help.
82240
We have $n$ distinct balls ($n&gt;7$) and want to randomly (and independently) distribute them into $N$ distinct boxes ($N&gt;n$) which are placed one next to the other. a) What is the probability that all balls are placed in consecutive boxes? b) What is the probability that all balls are placed in consecutive boxes and balls with number $1$, $4$ and $7$ are also placed in consecutive boxes?  I am not sure but I will try: a) There is a block of n balls which must be placed in one of the $N-n+1$ gaps formed if we arrange the boxes and fill $n$ consecutive of them. So there are $N-n+1$ places place this block. The $n$ balls in the block can be arranged in $n!$ ways. Also the N boxes can be arranged in $N!$ ways. So all in all we have $n!N!(N-n+1)$. b) Again we have $N!$ ways to arrange the boxes but the balls can be arranged in $(n-3)!$ ways, right? I don't know what is the total number of ways to place all balls in the boxes - plus that I see no restriction regarding the number of balls per box. Any help?
29199
Six identical yellow balls and four identical red balls are to be arranged in the circumference of a circle.  In how many ways it can be done?
68177
There are five boxes consisting of several identical coins. The first box contains 4 coins, the second box contains 9 coins, the third box contains 7 coins, the fourth box contains 10 coins, and the fifth box contains 3 coins. Some coins are taken from the boxes such that at least a coin taken from each box. In how many ways in taking the coins, such that the numbers of the coins taken are 12?  My attempt was solving it using repetition combination of which, there are only seven coins remaining considering that at least a coin taken from each box. By using the repetition combination, i.e. taking 7 from 9 (five boxes and four borders), resulting in 36 ways. However, there is no 36 ways in the choices.
114550
I was assigned the following problem for homework and would like to double check that I have approached it correctly.  Let $k \geq 2n$. In how many ways can we distribute $k$ sweets to $n$ children, if each child is supposed to get at least 2 of them?  This is a standard stars-and-bars problem, with the added constraint that each child gets at least two. In other words:  $x_1 + x_2 + \cdots + x_n = k$, such that $x_i \geq 2$, for $0 \leq i \leq n$  We can reduce this problem to solving over $y_i$ such that $y_i = x_i - 2$ and $y_i \geq 0$.  $\sum_{i=0}^n y_i = k$ or $\sum_{i=0}^n (x_i - 2) = k$ which reduces to  $\sum_{i=0}^n x_i = k + 2n$.  Plugging this into standard stars and bars we get $\binom{n + k + 2n - 1}{n - 1} = \binom{k + 3n - 1}{n - 1}$  Is this solution correct? Am I far off? This seems weird that the search space is bigger in the constrained version than the regular ($\geq 0$) version.
94802
I'm trying to find a way to calculate a problems such as this: if you have $n$ objects and $k$ indistinguishable boxes, how do you put in $n$ objects such that each box has no more than $C$, where $C \le k$.  For example, I have $6$ objects that need to be placed in $4$ boxes, where no box can have more than $4$ objects in it. I tried $\displaystyle\binom{n+k-1} {k-1}$. I got $84$ ways. But that is too big. When I do it by hand, I get $74$.   Would like actual explanation.
123141
What's the probability that a n-sided polygon made from n distinct random points on circle contain the centre?
112895
Let consider an urn model containing $n$ balls such that $m_g$ balls are green, $m_r$ are red and $m_b$ are blue and we have $n = m_g + m_r +m_b$.  Especially this means that the $m_g$ green balls are indistinguishable with respect to each other (and the same game for red and blue ones)  We draw $l \le n$ times without placing the balls back.   My question how many posibilities there exist for different $l$-draws taking the order and the indistinguishability of balls of same color into account.   More generally: the urn contains $m_1, m_2, ..., m_k$ balls with $\sum _i ^k m_i =n$ such that the $m_j$ are undistiguishable and one make $l$ draws without placing the balls back.   What is the cardinality of the resulting probability space?  Could anybody explain if there exist an intuitive way to describe and the count all cases?
49145
Make sequence of $9$ balls from 3 red, 3 blue, 3 green, in such a way that no two balls of the same colour are next to each other. In how many different ways can you do this? (symmetric arrangements must be counted only once).  To start with, all possible arrangements are $9! = 362880$.
87966
Let $x_1, \ldots ,x_n$ be distinct points in $\Bbb{R}^d$. Prove that there is a permutation $\sigma$ of $[n]$ such that the $\angle x_{\sigma(i−1)}x_{\sigma(i)}x_{\sigma(i+1)} &lt; \pi/2$ for any $i = 2,\ldots, n−1$.  So I have got this question, and I am not able to understand what &quot;a permutation $\sigma$ of $[n]$&quot; mean? Can someone explain to me this question and provide some hints to approach it? P.S. I have edited out some parts as it was a bit irrelevant.
95556
The following is a question I've made myself, but I need help in solving it:  Suppose there are 100 balls in a box. 20 balls are blue, 30 balls are green and 50 balls are yellow. Now we randomly pick out 10 balls out of the box (one ball after the other) and we don't put the balls back in the box.  What's the probability of the 10th ball being picked having color blue?  I tried thinking of all the possibilities of the the 10th ball being blue, divided by all of the possible combinations. I tried for several hours and couldn't figure out neither of them. Help is appreciated!
123788
Suppose that a box contains 10 red balls, 20 green balls, and 30 blue balls.   Suppose also that balls are drawn from the box one at a time at random.  What is the probability that all the red balls are drawn before the blue or green balls are themselves exhausted. What is the probability that as the last red ball is drawn, there remains at least one blue and one green left in the box.  The answer I was given is $\dfrac{7}{12}$ and a general equation is: $$ \dfrac{b g}{1-b}+\dfrac{b g}{1-g} $$ where $$ g=\dfrac{20}{60},b=\dfrac{30}{60} $$ but why?
19487
The number of different ways to choose $k$ out of $n$ unique items:   Without repetition and without order-significance: $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ Without repetition and with order-significance: $\binom{n}{k}\times k!=\frac{n!}{(n-k)!}$ With repetition and with order-significance: $n^k$ With repetition and without order-significance: $?$   Thank you!