New answers tagged

0

I've chosen to guide you along and not give you the answer directly, especially since you haven't said where you are stuck. I assume that you have some familiarity with binomial coefficients. A binomial coefficient $\binom{n}{k}$ is often read as "$n$ choose $k$", that is, "how many different ways can one pick $k$ out of a pool of $n$?" This applies to ...


0

By Bayes Theorem, we just need to compute the total probability the second card is a king, and then compute the fraction of that total that comes from the first card also being a king. The total probability the second card is a king is 4/52 * 3/51 + 48/52 * 4/51. The first summand is prob(first = king) * prob(second = king | first = king). The second ...


0

The conjecture is true if no non-empty proper subset of $\{a_1,\ldots,a_n\}$ sums to $0$. Hereafter I will make that assumption. Let $\langle b_0,\ldots,b_{n-1}\rangle$ be any permutation of the numbers $a_1,\ldots,a_n$. Extend it to an infinite periodic sequence $\langle b_k:k\in\Bbb N\rangle$ with period $n$ by setting $b_{\ell n+k}=b_k$ for ...


0

I think I see at least how to calculate this at all. here is my answer. Start with the tuple $$( \alpha_1, \cdots, \alpha_k )$$ Then form a set of the $k$ possible tuples by subtracting one from each entry. Note the size of this set. Continue making new sets by forming all possible tuples by subtracting one from each possible entry--every element of the ...


5

use the hockey stick identity. What we want is $\sum_{i=3}^{n}\binom{n}{3}6$ It is a known fact $\sum_{i=k}^n\binom{i}{k}=\binom{n+1}{k+1}$ Hence $\sum_{i=3}^{n}i(i-1)(i-2)=6\binom{n+1}{4}$


5

One way is to consider putting some/all/none of the chocolates on the left and the rest on the right. Each chocolate can go either way, so there are $2^n$ possibilities. But $2$ of these possibilities leave one side empty. Ignoring these, there are $2^n-2$ possibilities. Each pattern has a distinct reflection, swapping all those on the left with all ...


2

What are the odds that the second card drawn is a king? We pick the second card first, and there is a $\frac{1}{13}$ chance that it is a king. Any card will do for the first card. So the total probability is $1.$ What is the probability that both cards are kings? This probability is simply $\frac{1}{13} \times \frac{1}{17} = \frac{1}{221}.$ The conditional ...


2

You might as well pick them in the other order and ask the chance the second is a King given that the first is a King. Given that the first is a King, you are drawing from a $51$ card deck which has ????


1

EDIT: I misunderstood the question, corrected now. The probability of drawing a king after you have already drawn one king is $\frac{3}{51}=\frac{1}{17}$, since there are only 3 kings and 51 cards left.


1

Since $k=3$, there are $\binom31=1$ $1$-combinations: $\{1\}$, $\{2\}$, and $\{3\}$. There are $\binom32=3$ $2$-combinations: $\{1,2\}$, $\{1,3\}$, and $\{2,3\}$. And there is $\binom33=1$ $3$-combination, $\{1,2,3\}$. For $j=2$, for instance, $k-j-1=0$, so the $\Sigma^*$ expression expands to $$\begin{align*} ...


0

We have three heads H, followed by at least two tails T, which means we have three blocks HTT. We have no more heads, so the rest must be tails as well. There are 4 spots where we can put those Ts: x HTT x HTT x HTT x At every x there can be 0 or more Ts, for a total of 3 Ts. That can be done in three ways: 3 Ts at one spot, for a total of $4$ spots 2 Ts ...


3

To determine the number of solutions of the equation $$x_1 + x_2 + x_3 + x_4 = 20 \tag{1}$$ in the non-negative integers subject to the restrictions $x_1 < 7$ and $x_2 < 6$, we subtract the number of solutions in which $x_1 > 7$ or $x_2 > 5$ from the number of solutions of the equation. A particular solution of equation 1 corresponds to the ...


4

The classic way to do this is count the solutions without conditions, then use inclusion-exclusion to deal the cases which violate. So if $A$ is the set of all non-negative solutions to $x_1+x_2+x_3+x_4=20$, and $A_1$ is the set of such solutions with $x_1\geq 8$ and $A_2$ is the set of solutions with $x_2\geq 6$ then the set you seeking to count is ...


5

One way of looking at it is this: The number of ways is just the coefficient of $x^{20}$ in the expansion of $$(1+x+x^2+\cdots+x^7)(1+x+x^2+\cdots+x^5)(1+x+x^2+\cdots+x^{20})^{2}$$ Here, each factor represents a term in your equation (here I have written them in the same order). The powers of $x$ in the factor are the values the corresponding variable can ...


0

The important question here, is when is a date ambiguous? A date is ambiguous if the day (of the month) and the month (of the year) can be confused for one another; when the day can be a month as well. Since there are 12 months in a year, the day can only be confused with the month if the day is in $[1, 12]$, so there are twelve days each month where we ...


3

Consider (as example) point set of $86$ points: pts = { {-32548, 4941, 1746}, {-32182, -7188, -473}, {-31473, -484, -9790}, {-30188, -7755, 11126}, {-29817, 6257, 12734}, {-29196, 12267, -8910}, {-28393, -13023, -10705}, {-28249, 16917, 3300}, {-26936, 5415, -18260}, {-26336, -19736, 1372}, {-24723, -7559, ...


0

As an approximation, I would note that the AM-GM gives $$ \begin{align} 12376 &=\frac{n(n-1)\cdots(n-9)(n-10)}{11!}\\ &\lesssim\frac{(n-5)^{11}}{11!} \end{align} $$ Therefore, $$ n\gtrsim5+(11!\cdot12376)^{1/11}=16.5629 $$ Trying $n=17$ gives $\binom{17}{11}=12376$


-3

$a_n=2a_{n-1}+b_{n-1}$ $b_n=b_{n-1}+c_{n-1}+a_{n-1}$ $c_n=a_{n-1}+b_{n-1}$ Explanation: -First relation: $2a_{n-1}$ indicates sequences starting by 1... or 2... while the relation $a$ is for unconstrained sequences $b_{n-1}$ indicates 0... while $b$ is a sequence right-bounded by a $0$ -Second relation: $c_{n-1}$ indicates 01... while $c$ is any ...


0

$$(n-10)^{11}<n(n-1)\cdots(n-10)<n^{11}$$ so $$\root{11}\of{11!(12376)}<n<\root{11}\of{11!(12376)}+10$$ So that just gives you a few values of $n$ to try, and I'd recommend starting in the middle of that range.


0

You missed that there are six pairs of two sets, not four, then that you have to consider triplets of sets before you get to the intersection of all four.


1

You have used Inclusion-Exclusion, which is correct, but it goes further than that. Numbers with none of 3,4,5 have been subtracted three times in $|A_1|,|A_2|,|A_3|$, added back in three times in $|A_1\cap A_2|,|A_1\cap A_3|,|A_2\cap A_3|$, so must be subtracted again in $|A_1\cap A_2\cap A_3|$ Lastly, $|A_1\cap A_2\cap A_3\cap A_4|$ must be added back in.


1

I get $49$ of length $3$ by hand count. There are $11$ strings ending in $0$ because we subtract $x20$ and $200$. There are $14$ strings ending in each of $1$ or $3$ because to fail the first two characters must be $12$ or $20$. There are $10$ strings ending in $2$ because we subtract $x12, 122, 202$ Let $d(n)$ be the number of good $n$ character ...


0

It is good that you were alert. I believe everything you are doing is correct. If we continue, then $$\binom{n}{2} = \frac{\binom{20}{2}\binom{10}{2}}{20(10)} = \frac{171}{4}.$$ I think it is ok to have a decimal number here. This gives \begin{align*} \frac{n!}{2!(n-2)!} &= \frac{171}{4}\\ \implies \frac{n!}{(n-2)!} &= \frac{171}{2}\\ n(n-1) &= ...


2

You don't need to worry about order of the aces, nor identity of the other cards.   This is all about placement in the deck. In general, there are $\binom{n}{k}$ ways to select $k$ places from $n$. You have 4 aces and 52 places to put them in the deck.   Count the ways to select 4 of 52 places as the measure of the total space. For the ...


4

Let $S = \{1, 2, 3, \ldots, 3n - 2, 3n - 1, 3n\}$. Let \begin{align*} A & = \{k \in S \mid k \equiv 0 \pmod{3}\}\\ B & = \{k \in S \mid k \equiv 1 \pmod{3}\}\\ C & = \{k \in S \mid k \equiv 2 \pmod{3}\} \end{align*} Observe that $|A| = |B| = |C| = n$. We can choose three numbers from $S$ in the following ways: Choose $3$ elements of $A$, ...


1

Just like all of the "poker probability" questions, you may reword this as asking about the probability of a twenty-card hand satisfying the condition that it contains exactly three aces. Determine a useful sample space, preferably one which is equiprobable (one in which all outcomes are equally likely to occur). Determine the size of the sample space ...


1

Number of solutions to $a+b+c=k$ where $a,b,c$ are positive integers is $\binom{k-1}{2}$. So the number of ways to choose is $$ \sum_{k=1}^n\binom{3k-1}{2}={n\over2}(3n^2-1) $$ Edit: As JMoravitz notes below this counting accounts for ordered triples from the given set rather than subsets.


1

i believe the answer is ${10\choose 5}{5\choose 2}{3\choose 3}$ which can also be expressed as ${10\choose {5,2,3}}$. $3^{10}$ would be if there were no rules as to how many go in each room


0

Let $x_k$ denote the number of votes received by the $k$th candidate. Since one or more of the eleven eligible voters may not vote, the number of votes the candidates receive satisfies the inequality $$x_1 + x_2 + x_3 + x_4 + x_5 \leq 11 \tag{1}$$ Let $a$ denote the number of people who abstain from voting. Then $$x_1 + x_2 + x_3 + x_4 + x_5 + a = 11 ...


0

$a_n=a_{n-1}+2b_{n-1}$ $b_n=b_{n-1}+a_{n-1}$ $a_0=1,b_0=1,$ A note for blind delete-voters: This answer is nothing different fom the answer above except that it is a succint definition with two instead of 4 relations.


1

Here is a way to get started: Let $e_n$ be the number of words which start with a, let $b_n$ be the number of words which start with b, and let $c_n$ be the number of words which start with c. Then $a_n=e_n+b_n+c_n$ and $e_n=a_{n-1},\;\;$ $b_n=e_{n-1}+b_{n-1},\;\;$ and $c_n=e_{n-1}+c_{n-1}$.


3

Given a string of length $n-1$, you can add any of $0$, $1$, or $2$ to the end of it to get a string of length $n$, so you get $3 a(n-1)$ strings. However, if the last two elements of the original string were $01$, then adding $2$ is not allowed. How many such strings are there? Well, they are found by noting that after removing the $01$ you have an ...


0

Let S be the set of all solutions in nonnegative integers, and let $E_i$ be the set of solutions with $x_i\ge10$ for $1\le i\le4$. Using Inclusion-Exclusion, $\displaystyle|\overline{E_1}\cap\cdots\cap\overline{E_4}|=|S|-\sum_{i}|E_i|+\sum_{i<j}|E_i\cap E_j|-\sum_{i<j<k}|E_i\cap E_j\cap E_k|+|E_1\cap\cdots\cap E_4|$ $\hspace{1.0 ...


0

All solutions Split $(x_{1}+x_{2})+(x_{3}+x_{4})=21$. There are 16 different pairs $(x_{1}+x_{2},x_{3}+x_{4})$ (if you count $3+18$ different from $18+3$) simply $(3,18),(4,17),(5,16),...,(10,11),...,(18,3)$ For $(10,11)$ and $(11,10)$ you have $9$ combinations for $10$ and $8$ combinations for $11$. That makes $2\cdot 9 \cdot 8=144$ combinations. For ...


2

Write down $n-k$ "stars" to represent, in the abstract, books not taken. Then there are $n-k+1$ "gaps" between these stars, including the two endgaps. We must choose $k$ of these gaps to slip the $k$ books back into. The number of ways to do this is $\dbinom{n-k+1}{k}$. Remark: A nice recursion has been given by lulu. Alternately, one can verify that the ...


0

$k$ should be less than or equal to $\lceil n/2 \rceil$. with the above condition, the answer would be $\binom nk - (n-1)\binom {n-2}{k-2} + (n-2)\binom {n-3}{k-3}-...=\binom nk-\sum_{i=1}(-1)^i(n-i)\binom {n-i-1}{k-i-1}$ which is achieved using inclusion-exclusion principal. The first term is the number of ways of choosing k books out of n books (with no ...


1

Note: there is a simple recursion. If $F(k,n)$ is the answer you want, then we split the selections according to whether or not the first book in line is chosen. This immediately tells us that $$F(k,n)=F(k-1,n-2)+F(k,n-1)$$ Of course we also know that $F(k,n)=0$ if $k>\lceil \frac n2\rceil$, not to mention $F(0,n)=1$ and $F(1,n)=n$. Perhaps this ...


1

Note that $t$ does not appear in formula (E). The useful definition of $\binom{n}k$ here is $$\binom{n}k=\frac{n^{\underline k}}{k!}\;,$$ where $x^{\underline k}=\prod_{i=0}^{k-1}(x-i)$. Thus, $$\begin{align*} \sum_{j=0}^3(-1)^j\binom3j\binom{8-4j-1}2&=\binom30\binom72-\binom31\binom32+\binom32\binom{-1}2-\binom33\binom{-5}2\\ ...


2

This can be expressed as the coefficient of $x^{21}$ in the binomial expansion of $$(1+x+x^2+\cdots+x^9)^4$$ This is because the coefficient will represent the number of ways $4$ powers of $x$, ranging from $0$ to $9$, can add up to $21$. Thus you get the number of solutions to the given equation. Each variable has been replaced with a series in $x$ whose ...


1

To solve the problem using permutations: Note that there are $5!$ ways to arrange the letters if you ignore the condition. Transposing $A,D$ shows that there are exactly as many permutations with $A$ first as there are with $D$ first, so the answer is $\frac {5!}2$.


2

It is precisely because we know that $A$ comes before $D$ that we want $\binom52$. The point is that once we know which two positions in the string are occupied by $A$ and $D$, we automatically know which one contains $A$ and which one contains $D$: the earlier one must contain $A$. There are $\binom52$ ways to choose two positions out of the five in the ...


1

Counting down from all $11$ voting to none voting, without any other restrictions, stars and bars will give $\binom{15}4 + \binom{14}4 + .... + \binom44$ ways To take into account that no one gets more than $5$ votes, we "pre-place" $6$ votes for any of the five candidates, and subtract all such cases, viz. $5\left[\binom94 + \binom84 + .... ...


0

Assuming I understand what you mean, it doesn't matter $who$ votes, just the number of votes is important (i.e. they are identical). The total number of ways to distribute 11 votes among 5 candidates is $\binom{15}{4}$, from this you need to subtract the number of ways a candidate can get more than 5 votes. Only 1 out of 5 such candidates is possible, ...


0

For any of the three solutions, for each die, there is a $\tfrac{1}{6}$ chance of rolling the right number, so there is a $\tfrac{1}{6^{10}}$ chance of that specific combination turning up. But it doesn't matter if the first die rolled a $6$ and the last the $3$ you needed, or the other way around. So we're going to try to figure out how many possible ways ...


2

Given any $40$ people, at least four of them were born in the same month of the year. The the phrase "at least four of them were born in the same month of the year" means just that there is some subset of four people out of the $40$ who share a common birth month. It does not mean that everyone belongs to some clique of four people with the same birth ...


5

The given sum is $$ \sum_{k=0}^{n-1}\binom{n}{k}\left(-{1\over2}\right)^k=\sum_{k=0}^{n}\binom{n}{k}\left(-{1\over2}\right)^k-\left(-{1\over2}\right)^n=\left(1-{1\over2}\right)^n-\left(-{1\over2}\right)^n\\=\left({1\over2}\right)^n-\left(-{1\over2}\right)^n $$


1

Try to use the binomial formula on $(1+(-\frac{1}{2}))^n$ This will give you to the proof of the statement.


1

Hint: Suppose otherwise. Then at most three people are born in each of the twelve months of the year. What can you conclude?


-1

Equality holds for $x=1$. Then, taking the derivative, $$n\le n(1+x)^{n-1}.$$


1

I guess an easy way to do it would be induction if you are having trouble with binomial theorem. For base case n=1 it is trivial $$1+x \le 1+x$$ Let it be true for some $n=k$. We have $$\begin{align} 1+kx \le(1+x)^k\\ \Rightarrow (1+kx)(1+x) \le (1+x)^{k+1}\\ \Rightarrow 1+x(k+1)+kx^2\le(1+x)^{k+1}\\ \end{align}$$ But if $k\gt 0$ and $x\gt0$ we must have ...



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