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0

Combine the 40w and 65w bulbs into one group (Low) of nine bulbs. The other group (High) has the six 75w bulbs. Let $X$ be the number of H bulbs you get in two draws. The random variable can take values 0, 1, and 2. $P(X = i) = {6 \choose i}{9 \choose {2-i}}/{15 \choose 2},$ for $i = 0,1,2.$ Find all three and check that they add to 1. You want the ...


1

Imagine if you will that each and every bulb has a unique id label. I.e. you have 40watt bulbs: $A_1, A_2, A_3,A_4$, you have 65watt bulbs $B_1,\dots, B_5$, you have 75watt bulbs $C_1,\dots, C_6$. Notice that every pair of two bulbs is equally likely to be drawn. For convenience, let us assume order doesn't matter. (The same arguments will apply for if ...


2

You are correct. Here' sanother way to think about it: Let the boy and girl together be a "pair". The pair can be in position (1,2) or (2,3) or ... or (7,8): 7 possibilities. The pair can be boy first or girl first - so multiply by 2. The "boy" and "girl slots are now determined by the position of the boy nad the girl from the first pair. The other three ...


0

This is the same problem as: How many ways are there to rearrange the letters of the word $BANANA$ Have the kids stand in a line ordered alphabetically (or by shoesize or whiteness of teeth, it really doesn't matter). For each arrangement of the letters in the word banana, say for example $BAAANN$, give the peanutbutter cookie to the child standing in ...


1

In 1. if order matters, we have both distinguishable people and distinguishable chairs, so it's simply leaving one chair out (there are $10$ available) and then ordering the people, i.e. $10 \cdot 9! = 10!$. The other way would be to number the chairs from $1$ to $10$ and have person $i$ sit on chair $i$, giving $10!$ directly. If order didn't matter in ...


1

There is an old joke about the ways that people in different professional fields "prove" that all odd numbers greater than $2$ are prime. One way is, "$3$ is a prime, $5$ is a prime, $7$ is a prime, therefore all odd numbers greater than $2$ are prime." You do not want people thinking your reasoning was like that. There are a few ways to show that there is ...


0

Sorry i was being silly calculating $\mathbb{E}(X)=\sum_{k} \mathbb{P}(X=k)$ rather than $\mathbb{E}(X)=\sum_{k} k\mathbb{P}(X=k)$ i blame it on a long day!! I want to just run through the rest of their solution as they miss a lot of detail, they don't explain why the expectation divided by $65$ gives the answer. Here is my extended solution which aims to ...


0

Let A(x) = 1 + x + x^2/2 + 2 x^3/3! + 6 x^4/4! and B(x) = x + x^2/2 + 2 x^3/3! We want the coefficient of x^12/12! in A(B(x)) which is 1478400. As an added bonus we know the number of ways to seat 0,1,2,...,12 people is: 1, 1, 2, 7, 29, 150, 820, 4130, 19670, 80080, 285600, 739200, 1478400.


0

The expected number of pairs is $$\frac{1}{365} \cdot \frac{n(n-1)}{2}$$ However, this also includes cases that there are two, three, four, or even more pairs. Therefore the number is a little lower.


2

Consider your selected numbers $\{a_1, a_2, \ldots a_{14}\} \bmod 13$. Then they must each be in a residue class, $\{r_1, r_2, \ldots r_{14}\} $ - but there are only $13$ residue classes, so at least $2$ must be in the same class. The difference of any $2$ numbers in the same residue class is divisible by $13$. Note that by using prime powers different to ...


1

Intuitively, you could think in terms of modular arithmetic: Call 3 members of your set $a,b$ and $c$. Suppose $a-b=k_1 \mod 13$, and $a-c= k_2 \mod 13$. Now if $k_1=k_2$, then $b-c$ will be divisible by 13, so you need $k_1 \neq k_2$ to have problems. But you have 13 such equations, and only 12 different values for $k$, so apply the pigeonhole principle ...


2

You can break up the set of integers in to $13$ sets which do not intersect. There are the multiples of of $13$, numbers whose remainder when divided by $13$ is $1$, numbers whose remainder is $2$ and so on. If you pick $14$ distinct numbers, at least two of them have to belong to set same set by the pigeonhole principle. The difference of these two will be ...


0

It depends on what you mean by "teams". If your "teams" are distinguishable then there are 8!/2^4. If your "teams" are generic indistinguishable groups then you still need to divide by 4! One way to see this is to imagine 4 empty glasses in a row. Put the integers 1,2,3,4,5,6,7,8 into a pitcher and stir them up really good. Pour two integers into each ...


4

We assume the teams are not labelled. Line up the $8$ people in a row, in order of age, or weight, or student number. The leftmost person has $7$ ways to choose her team mate. For every such choice, the leftmost person not yet teamed up has $5$ ways to choose her team mate. And for every such choice, the leftmost person not yet teamed up has $3$ choices, ...


0

Simply represent as $abc\vee\vee\vee$ where each $\vee$ is a 'slot' where one of $x,y,z$ may be placed, although $x$ and $y$ must occupy the first $2$ slots. So $a\vee bc\vee \vee$, $a\vee b\vee \vee c$, etc. possibilities. $a,b,c$ remain in same order. Number of permutations is $P(6,3)={{6!}\over {3!}}$. However the $\vee$'s cannot be distinguished so ...


0

A combinatorial interpretation: you are correct in thinking that the sum is zero. One way to express this is that the number of subsets of odd cardinality of an $n$ element set is the same as the number of subsets of even cardinality of an $n$ element set. Indeed we can put these in bijection by taking the symmetric difference of a set with $\{1\}$, i.e. $$ ...


3

The first team can be selected in $\binom{8}{2}$ ways, the second team can be selected in $\binom{6}{2}$ ways, and the third team can be formed in $\binom{4}{2}$ ways, and the remaining $2$ form the last team of $4$. Thus there are: $\binom{8}{2}\binom{6}{2}\binom{4}{2}$ ways.


0

$8!$ overcounts more than just $a,b$ and $b,a$, it also accounts for the order of things and in your question order does not matter. I believe the answer is ${8\choose2}\cdot{6\choose 2}\cdot{4\choose 2}$?


0

Before like terms are collected, a typical term of the product has an exponent of the form $$a_0\cdot 10\cdot0+a_1\cdot10\cdot 1+\ldots+a_{99}\cdot10\cdot99=\sum_{k=0}^{99}10ka_k\;,$$ where each $a_k\in\{0,1,2,3,4,5,6,7,8,9\}$; we can take this to be the weight of one element of the set in question. There are many sets that could be made to work with a ...


2

To avoid argumentation over the set of powers of $3$ (of which you only test the first few members), you could simply show: $X=\emptyset$ certainly does not work. And if $X\ne \emptyset$, say $a\in X$, then the number of maps $X\to \{1,2,3\}$ is divisible by $3$ cause it is three times the number of maps $(X\setminus\{a\})\to\{1,2,3\}$. The claim then ...


4

Any symmetry operations of the trirectangular tetrahedron will send vertices to vertices, edges to edges and faces to faces. Since the vertex with three right angles is distinguished, any symmetry operation will leave that vertex unchanged. If the lengths of the edges attached to that vertex are all different, any symmetry operation will leave them ...


3

The general idea is correct, but: a minor issue, but important: you say that $|z|^{|x|}$ is a set of functions. That is incorrect. $z^x$ is the set of functions $f\colon x\to z$, but $|z|^{|x|}$ is the number of all functions $f\colon x\to z$. Major issue: Your conclusion that $3^n\ne 1000$ just from observing $3^1$, $3^2$, and $3^3$ is completely and ...


5

Your argument that $3^n\neq 1000$ seems weird and not complete. But the basic idea is good.


2

I assume that $(abc)(def)(ghi)(jkl)$ is considered the same result as $(def)(ghi)(jkl)(abc)$ (i.e. rotations of tables is irrelevant) and $(bca)(def)(igh)(klj)$ (i.e. rotations within tables is irrelevant), but is considered different from $(bac)(def)(ghi)(jkl)$ (mirrors of tables considered different) and is considered different from $(def)(abc)(ghi)(jkl)$ ...


0

If you replace the sets $C_j$ by their complements $C_j'$, then each $B_i$ must be contained in some $C_j'$. This is called a covering design. The La Jolla Covering Repository is a great place to look.


1

Let $\displaystyle P(x)=\sum_{r=0}^n (-1)^r\binom{n}{r}(x-r)^n$; we want to show that $P(x)=n!$. Since $\displaystyle P(x)=\sum_{r=0}^n(-1)^r\binom{n}{r}\bigg[\sum_{j=0}^n\binom{n}{j}(-r)^jx^{n-j}\bigg]=\sum_{j=0}^n (-1)^j\binom{n}{j}\bigg[\sum_{r=0}^n(-1)^r\binom{n}{r}r^j\bigg]x^{n-j}$, it suffices to show that ...


3

You can make your method work. If $x$ is in the last position, then there are $5!$ ways to arrange the other letters. If $x$ is in the fifth position, then there are $3$ ways to select the letter in the last position from among $a, b, c$ and $4!$ ways to arrange the letters before $x$. Therefore, there are $3 \cdot 4!$ arrangements with $x$ in the fourth ...


3

We are given a poset $\{a,b,c,x,y,z\}$ where the only comparisons defined are $x>y$ and $x>z$. We try to find how many ways there are to extend this to a chain. First: either $x>y>z$ or $x>z>y$. Two choices. Without loss of generality, assume it is the first one. Now, decide where to place $a$ in that list. Either $a>x>y>z$ ...


5

I will assume that after is not restricted to immediately after. There are $6!$ ways that the people can be arranged without restriction. There are $3!$ relative orders for $x,y,z$. Exactly $2$ of these relative orders have $x$ after $y$ and $z$. So the number of ways is $6!\times \frac{2}{3!}$. We can do the calculation more slowly. Imagine $6$ chairs in ...


1

We have $\binom{8}{2}$ choices for Match 1, $\binom{6}{2}$ choices for Match 2, $\binom{4}{2}$ choices for Match 3, and $\binom{2}{2}$ choices for Match 4. So for 8 teams, there are $$ \binom{8}{2}\times\binom{6}{2}\times \binom{4}{2}\times\binom{2}{2} $$ possibilities. Can you generalize this?


0

The symmetry of this problem provides another way to look at it: you choose a coin at random and look at one of its faces at random. That face shows something (maybe heads, maybe tails). First question: What is the probability that the other face of the coin is the same as the face you can see? The answer to that question is clearly $\frac23$, since two ...


0

It seems to me you can use induction and the following transformation on sequences $(a_1,\ldots,a_n)$ of positive integers with $a_i\le i$ and even sum: $$ (a_1,\ldots,a_{n-1},a_n)\mapsto \begin{cases} (a_1,\ldots,a_{n-2})&\text{if}\quad a_{n-1}=a_n,\\ (a_1,\ldots,a_{n-2},|a_n-a_{n-1}|)&\text{if}\quad a_{n-1}\not=a_n. \end{cases} $$ The resulting ...


0

Regarding the question "What are the total number of ways in which a specified number is visible on both the dice?" I am assuming that a particular number (i.e. 1) if visible on the top face after the throw is distinguishable from the case where the 1 is visible on one of the side faces after the throw. If this is the case then one die can be viewed in 8 ...


0

The number of ways, n, to put 12 pennies into 3 bins so that each bin has an odd number of pennies is the coefficient of x^n in the expansion of ( x/(1-x) )^3. Which is 0. nn = 15; CoefficientList[Series[(x/(1 - x^2))^3, {x, 0, nn}], x] {0, 0, 0, 1, 0, 3, 0, 6, 0, 10, 0, 15, 0, 21, 0, 28}.


0

Let $a=1+x^2+x^4+x^6+x^8+\cdots$ If we multiply this by $x^2$, $$ax^2=x^2(1+x^2+x^4+x^6+x^8+\cdots)=x^2+x^4+x^6+x^8+x^{10}+\cdots$$ Notice that this is $1$ less than our original value of $a$. $$ax^2=a-1$$ Simplifying.... $$ax^2-a=-1$$ $$a(x^2-1)=-1$$ $$a=\frac{-1}{x^2-1}=\frac{1}{1-x^2}$$ This is often not the most common method, but it is still quite ...


1

Just noticed that it's impossible, and so must be a misprint. The sum of three odd numbers must be odd ($(2p+1)+(2q+1)+(2r+1)=2(p+q+r+1)+1, \forall p,q,r \in Z$) so no three odd bins can add to 12. I see Travis's suggestion of substitution would work in general.


0

Hint $$1 + u + u^2 + u^3 + \cdots = \frac{1}{1 - u}.$$


2

Let me suppose that you want $$A(x)=\sum_{n=0}^\infty a_i x^i$$ So, let us mutliply both sides by the denominator $$8+14x-50x^2=(1-7x^2+6x^3)\sum_{n=0}^\infty a_i x^i$$ Now, decompose the product $$8+14x-50x^2=\sum_{n=0}^\infty a_i x^i-7\sum_{n=0}^\infty a_i x^{i+2}+6\sum_{n=0}^\infty a_i x^{i+3}$$ So, for the constant term $$8=a_0$$ For the first power of ...


-1

a) The given that all dishes are different, then $$ N_{orders}=15^8 $$ b) In this case, we can calculate how many different orders for each type and then combine them: $$ N_{orders} = 7^3 \times 4^3 \times 4^2 $$


1

a) First, we note that repetition is allowed, and the order in which we order the dishes is unimportant. Therefore, we use the formula ${k+n-1 \choose k}$. In this case, $n=15$ and $k=8$. So, the number of possible combinations of dishes is $${8+15-1 \choose 8} = {22 \choose 8} = 319770.$$ b) We will need to apply the above formula three times. For the ...


1

HINT: You need either to say what $\Phi_{P_n}$ is or include a link to your earlier question; I added the missing definition to your question. The most straightforward way to show that ${\Phi}^*_{P_n}=\Phi_{P_n}$ is to show that for each $n$ and $k$, the number of partitions of $n$ with $k$ parts is the same as the number of partitions of $n$ with largest ...


0

Adayah’s answer is the most straightforward. You can also prove it by induction on $n$. Suppose that for each $k\le n$ you know that it takes exactly $k-1$ iterations. Now start with a set of $n+1$ elements. After the first iteration you will have two sets, say $K$ and $L$, of cardinalities $k$ and $\ell$, respectively. Clearly $k+\ell=n+1$ and $k,\ell\ge ...


2

HINT: Since the company opens at most one store in a block, if a town has $n$ blocks, we can represent any possible way of opening stores by a string of $n$ zeroes and ones: a $0$ represents a block in which the company does not open a store, and a $1$ represents a block in which the company does open a store. The other restriction means that we want only ...


0

Consider leftmost part: there must be either store, then gap, then something or gap then something. So $f(n)=f(n-2)+f(n-1)$.


3

Suppose you have $n$ elements $A, B, C...$ and $m$ of each. Then the length of the string has to be $nm$. So the number of permutations, i.e. strings, is $$\frac{(nm)!}{(m!)^n}$$


2

You can do the calculation on the recursive formula for that problem. You can observe the following: we don't need to restrict ourselfs to regular polygons, the numbers remain the same, if strictly convex polygons are used. there are $k \cdot 2^{k-3}$ ways to combine $k$ nodes to paths such that no line on the path crosses another line of the path (choose ...


2

If you can trust my sage calculations, the circulant on 15 vertices with connection set $\{\pm3,\pm5\}$ is not isomorphic to any circulant with connection set $\{\pm1,\pm i\}$ for $i=2,\ldots,7$. (I believe your assumption may be true for circulants of prime power order.)


1

using the nested sums the general case can be reduced to the $k=0$ case with an interesting modification ... $$ \begin{align} &\sum_{r=0}^{n} { \binom{n}{r} (-1)^r(k-r)^n } \\ \\=&\sum_{r=0}^{n} { \binom{n}{r} (-1)^r \sum_{i=0}^{n} { \binom{n}{i} k^i(-r)^{n-i} }}\\ \\=&\sum_{i=0}^{n} {\binom{n}{i} k^i \sum_{r=0}^{n} { \binom{n}{r} (-1)^r ...


2

Choose three digits from $12345678$. So ${8 \choose 3} = \frac{8!}{5!3!} = 56$


1

Since the three digits must be distinct, we must select three of the eight elements in the set $S = \{1, 2, 3, 4, 5, 6, 7, 8\}$. Once we have selected these digits, there is only one way to order them so that the hundreds digit is less than the tens digit, which, in turn, is less than the units digit. Thus, the number of ways we can construct a three-digit ...



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