In a race there are $n$ horses. In that race, more than one horse may get the same position. For example, a race with 2 horses can finish in 3 ways:   Both first Horse 1 first and horse 2 second Horse 2 first and horse 1 second   How can I find out the number of ways the race can finish for any $n$. Is there any recurrence relation to calculate the number of ways? I need help to understand the solving way with better explanation.
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In a race there are n horses.In the race more than one horse may get the same position. For example, 2 horses can finish in 3 ways.   Both first horse1 first and horse2 second horse2 first and horse1 second   In how many ways,the race can finish so that a particular horse can never be first.If f(n) is the number of ways, is there any recurrence relation to get f(n).
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There are N plant pots available in nursery placed in a straight line. Bob is planning to take few of these pots. But whatever number of pots he is going to take, he will take no two successive pots.(i.e. if he is  taking a pot i from nursery, he will not take i-1 and i+1 pots.)  So given N, we need to calculate how many different set of pots he could select. He need to take at least one pot from nursery.  Example : Let N=3 then answer is 4. Please help to solve this problem
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This is a question that has received various answers. This questions seems simple, but it proves to be rather tricky.  So :  We flip a coin (H-T) independently $10$ times. Let's for the moment not care about whether it is fair or not.   How many $3$-head sequences are there?  Solution:  Some say that the answer is $\displaystyle\frac{10!}{7!3!}$, but this is the number of ways  by which $3$ heads can be arranged in $10$ positions. This answer , however, does not say anything about the remaining $7$ positions.   Each of these seven positions can be H or T. The total number of $7$-element sequences is $1\cdot1\cdot1\cdot2^7=2^7$. Then we have to multiply this by $\displaystyle\frac{10!}{7!3!}$ as the three heads can be anywhere. But I have found that this is not the correct answer. The correct is just $\displaystyle\frac{10!}{7!3!}$. So why not multiply by $2^7$?
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How to use the  method?  Say I want to find number of combinations I can get with $x_1+x_2+x_3+x_4=22$,  where $x_i\in\mathbb{N}$.  Is this the correct time to apply the method?  This is being repurposed in an effort to cut down on duplicates, see here:     and here:
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If a set has $n$ elements, how many transitive relations are there on it?  For example if set $A$ has $2$ elements then how many transitive relations. I know the total number of relations is $16$ but how to find only the transitive relations? Is there a formula for finding this or is it a counting problem?  Also how to find this for any number of elements $n$?
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This question has probably been asked before, but just to be clear here, I am NOT asking for the answer, I know the answer.  What i want to know is why my solution is not equivalent to the actual solution.  Three of a kind in a deck means you draw 3 cards of the same rank, and 2 other cards each of different ranks.  So. The solution in my book is the following: (13 C 1)(4 C 3) * (12 C 2)(4 C 1)(4 C 1)  The logic is the following:  1) Choose 1 out of the 13 ranks, then for the 4 cards in the chosen rank, pick 3.  2) Choose a combination of 2 ranks from the remaining 12, then for the 4 cards in each rank, choose 1.  Total combinations: 54912  My solution is the following:  ((13 C 1)(4 C 3)) * ((12 C 1)(4 C 1)) * ((11 C 1)(4 C 1))  1) My logic for the 3 of a kind is the same as the solution.   2) Choose 1 of the remaining 12 ranks, and 1 card from the 4 cards in the rank.  3) Choose 1 of the remaining 11 ranks, and 1 card from the 4 cards in the rank.  Total combinations: 109824  My solution says that the combinations are exactly double of the book's solution.  I'm just very confused as to why this is. If anyone could explain this to me I would be very appreciative, thank you.
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Say, we have 4 red balls, 4 blue balls, 4 green balls, 4 yellow balls. How many ways are there to form a sequence of 10 balls such that every color of ball occurs at least twice?  My thought: r = red b = blue g = green y = yellow * = undetermined  Every arrangments should in this form: rrbbggyy**  Case 1: two * are same  $\frac{\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1}10!}{4!2!2!2!}$  Case 2: two * are different  $\frac{\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}10!}{3!3!2!2!}$  Result = Case 1 + Case 2  Any suggestions would be appreciated!
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Edit 3. Fixed question to be more clear and include current solution Problem Two players player 1 and player 2 plays a game of fair coin flipping. Player 1 starts with $A$ coins and Player 2 with $B$ coins. Let $P_n$ defines the probability of player 1 winning all coins in a game when player 1 has $n$ coins. Find the recurrence relation of $P_n$ and then find its closed formula. This formula can then be used to know the probability of player 1 winning if he has, i.e. twice the coins player 2 has. Solution? I know that $P_0 = 0$, because if she has $0$ coins, then she can't win, what about $P_{A+B}?$ $$P_n = 0.5P_{n+1} + 0.5P_{n-1}$$ Which gives the characteristic equation: $$x^2-2x+1 = (x-1)^2=0 \implies x=1\ (double\ root)$$ This can then be put into the general solution for linear homogenous recurrence relations: $P_{hn} = C_1+nC_2$ and here I'm stuck once again. To find these two values I'd need two boundary points
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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!
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In how many ways can 3 men and their wives be made stand in a line such that none of the 3 men stand in a position that is ahead of his wife?  What is the best way to tackle such problems?   Here is how I am tackling this problem      M1 M2 M3  -&gt; Men     W1 W2 W3  -&gt; Wives      a.      1  2   3   4    5    6      W1 M1  W2  M2   W3   M3     -&gt;         With the above in mind, W1 M1 is a pair and W2 M2 is a pair, W3 M3 is a pair.     so, there can be 3 interchanges - if we assume that W1 M1 stick to each other.      b.        W1 W2  M1  M2   [W3   M3]  W1 W2 can interchange in 2 ways.  M1 M2 can interchange in 2 ways   4 ways..* 2 ( when W3 M3 will be at first) = 8 ways.          c.  W1 W2  W3  M1   M2   M3      W1 W2 W3 can be arranged in 3 ways     M1 M2 M3 can be arranged in 3 ways.     total ways are 9 ways.       -----------------      so adding all -  20 ways.   However, the answer is not 20 ways. What is wrong in my thought process here?
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Assume that there are $n$ people line up to get into $m$ different clubs. How many ways to do it if each people is different and order of people in the line matters.  Sol. Clearly, each people has $m$ different choices for clubs. So there are $m^n$ ways of choosing clubs. Now we count number of different lines. Let $x_i$ be the number of people lining to get to the club $i$, $1 \leq i \leq m$, $x_i \geq 0$.  Then $$x_1 + x_2 + ... + x_m = n$$ with $x_i$ non-negative integers. So there are ${m+n-1}\choose{n-1}$ ways, ignore order in each line. So there are ${m+n-1}\choose{n-1}$$ x_1!x_2!...x_n!$ possible lines.  The answer is weird since it is in $x_1, ...,x_n$ terms, and I think it is wrong. However, I am not sure how to fix it.  Any help please ?
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Let's say $A_n$ is the number of binary string that has length $n$ and does not contain the substring 101. Calculate $A_n$ for $n=1,2\cdots8.$ Find a recurrence relation for $A_n$. What does the solution of that recurrence look like?  These are the solutions that I have found for calculations for $A_n$. $1, 4, 7, 12, 20, 32, 48, 96$.  I've calculated this by hand. But how I do find the recurrence? I see that 4, 7, 12, 30 are Fibonacci - 1, but not after or before that.  But I'm not sure how to do this or if this is even correct.
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So I had asked a question prior to this one about recurrence relations, but apparently it was a bad one to ask. So I'm trying again to understand how to solve these babies... Here it is:  $$ 3a_{n+1}-4a_n=0, n&gt;=0, a_1=5 $$  What is the general process for solving a relation like this?
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I've seen a few similar threads to this on different forums but they don't seem to conclude to a satisfactory answer. My question is this:  If you have 3 horses, A, B, and C and you know the winning probabilities of each horse racing against each other as a pair, how do you work out their winning probabilities if all 3 horses race together? Is there a formula for n number of horses?  So you know the probabilities of: A beating B A beating C BA BC CA CB  I initially thought that: p(A winning) = p(A beat B). p(A beat C) But this is clearly wrong.  Can someone please help!! Many thanks Matt
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I am not able to solve the above recurrence relation which involves $n$. The initial conditions $a_0 = 1, a_1 = 2, a_2 = 3$, for $n \geq 0$. Please try to give closed form in terms of arbitrary $a_i \geq 1 $ for $i \geq 0$
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Omg this is driving me crazy seriously, it's a subproblem for a bigger problem, and i'm stuck on it.  Anyways i need the number of ways to pick $x[1]$ ammount of objects type $1$, $x[2]$ ammount of objects type $2$, $x[3]$ ammounts of objects type $3$ etc etc such that $$x[1] + x[2] + \ldots x[n] = k\;.$$ Order of the objects of course doesn't matter.  I know there is a simple formula for this something like $\binom{k + n}n$ or something like that i just can't find it anywhere on google.
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A coin is tossed $n$ times. What is the probability of getting odd number of heads?   I started this chapter sometimes ago and faced in front of a tough problem. At first I started considering cases. Case-I : The probability of getting 1 head.Case-II : The probability of getting 3 head and so on. But there are many cases. So how can I solved this . Please help me. Thank you!