heureka

We have 10 standard 6-sided dice, all different colors.

In how many ways can we roll them to get a sum of 20?

1.

The number of ways of partitioning 20 into exactly 10 distinct parts with  the 6 numbers {1,2,3,4,5,6} we have:

30 partitions of 20 into ten distinct parts:

The 30 distinct partitions with the sum 20 are:

   1.) {2,2,2,2,2,2,2,2,2,2}
   2.) {1,2,2,2,2,2,2,2,2,3}
   3.) {1,1,2,2,2,2,2,2,3,3}
   4.) {1,1,2,2,2,2,2,2,2,4}
   5.) {1,1,1,2,2,2,2,3,3,3} 
   6.) {1,1,1,2,2,2,2,2,3,4}
   7.) {1,1,1,2,2,2,2,2,2,5}
   8.) {1,1,1,1,2,2,3,3,3,3}
   9.) {1,1,1,1,2,2,2,3,3,4}
  10.) {1,1,1,1,2,2,2,2,4,4}
  11.) {1,1,1,1,2,2,2,2,3,5}
  12.) {1,1,1,1,2,2,2,2,2,6}
  13.) {1,1,1,1,1,3,3,3,3,3}
  14.) {1,1,1,1,1,2,3,3,3,4}
  15.) {1,1,1,1,1,2,2,3,4,4}
  16.) {1,1,1,1,1,2,2,3,3,5}
  17.) {1,1,1,1,1,2,2,2,4,5}
  18.) {1,1,1,1,1,2,2,2,3,6}
  19.) {1,1,1,1,1,1,3,3,4,4}
  20.) {1,1,1,1,1,1,3,3,3,5}
  21.) {1,1,1,1,1,1,2,4,4,4}
  22.) {1,1,1,1,1,1,2,3,4,5}
  23.) {1,1,1,1,1,1,2,3,3,6}
  24.) {1,1,1,1,1,1,2,2,5,5}
  25.) {1,1,1,1,1,1,2,2,4,6}
  26.) {1,1,1,1,1,1,1,4,4,5}
  27.) {1,1,1,1,1,1,1,3,5,5}
  28.) {1,1,1,1,1,1,1,3,4,6}
  29.) {1,1,1,1,1,1,1,2,5,6}
  30.) {1,1,1,1,1,1,1,1,6,6}

2.

In how many ways can we roll them to get a sum of 20?
We need the permutations of each partitions and the sum is the answer.

$\begin{array}{rcll} 1.)& \{2,2,2,2,2,2,2,2,2,2\} & \frac{10!}{10!} &=& 1 \\ 2.)& \{1,2,2,2,2,2,2,2,2,3\} & \frac{10!}{1!8!1!} &=& 90 \\ 3.)& \{1,1,2,2,2,2,2,2,3,3\} & \frac{10!}{2!6!2!} &=& 1260 \\ 4.)& \{1,1,2,2,2,2,2,2,2,4\} & \frac{10!}{2!6!1!} &=& 360 \\ 5.)& \{1,1,1,2,2,2,2,3,3,3\} & \frac{10!}{3!4!3!} &=& 4200 \\ 6.)& \{1,1,1,2,2,2,2,2,3,4\} & \frac{10!}{3!5!1!1!} &=& 5040 \\ 7.)& \{1,1,1,2,2,2,2,2,2,5\} & \frac{10!}{3!6!1!} &=& 840 \\ 8.)& \{1,1,1,1,2,2,3,3,3,3\} & \frac{10!}{4!2!4!} &=& 3150 \\ 9.)& \{1,1,1,1,2,2,2,3,3,4\} & \frac{10!}{4!3!2!1!} &=& 12600 \\ 10.)& \{1,1,1,1,2,2,2,2,4,4\} & \frac{10!}{4!4!2!} &=& 3150 \\ 11.)& \{1,1,1,1,2,2,2,2,3,5\} & \frac{10!}{4!4!1!1!} &=& 6300 \\ 12.)& \{1,1,1,1,2,2,2,2,2,6\} & \frac{10!}{4!5!1!} &=& 1260 \\ 13.)& \{1,1,1,1,1,3,3,3,3,3\} & \frac{10!}{5!5!} &=& 252 \\ 14.)& \{1,1,1,1,1,2,3,3,3,4\} & \frac{10!}{5!1!3!1!} &=& 5040 \\ 15.)& \{1,1,1,1,1,2,2,3,4,4\} & \frac{10!}{5!2!1!2!} &=& 7560 \\ 16.)& \{1,1,1,1,1,2,2,3,3,5\} & \frac{10!}{5!2!2!1!} &=& 7560 \\ 17.)& \{1,1,1,1,1,2,2,2,4,5\} & \frac{10!}{5!3!1!1!} &=& 5040 \\ 18.)& \{1,1,1,1,1,2,2,2,3,6\} & \frac{10!}{5!3!1!1!} &=& 5040 \\ 19.)& \{1,1,1,1,1,1,3,3,4,4\} & \frac{10!}{6!2!2!} &=& 1260 \\ 20.)& \{1,1,1,1,1,1,3,3,3,5\} & \frac{10!}{6!3!1!} &=& 840 \\ 21.)& \{1,1,1,1,1,1,2,4,4,4\} & \frac{10!}{6!1!3!} &=& 840 \\ 22.)& \{1,1,1,1,1,1,2,3,4,5\} & \frac{10!}{6!1!1!1!1!} &=& 5040 \\ 23.)& \{1,1,1,1,1,1,2,3,3,6\} & \frac{10!}{6!1!2!1!} &=& 2520 \\ 24.)& \{1,1,1,1,1,1,2,2,5,5\} & \frac{10!}{6!2!2!} &=& 1260 \\ 25.)& \{1,1,1,1,1,1,2,2,4,6\} & \frac{10!}{6!2!1!1!} &=& 2520 \\ 26.)& \{1,1,1,1,1,1,1,4,4,5\} & \frac{10!}{7!2!1!} &=& 360 \\ 27.)& \{1,1,1,1,1,1,1,3,5,5\} & \frac{10!}{7!1!2!} &=& 360 \\ 28.)& \{1,1,1,1,1,1,1,3,4,6\} & \frac{10!}{7!1!1!1!} &=& 720 \\ 29.)& \{1,1,1,1,1,1,1,2,5,6\} & \frac{10!}{7!1!1!1!} &=& 720 \\ 30.)& \{1,1,1,1,1,1,1,1,6,6\} & \frac{10!}{8!2!} &=& 45 \\ \hline && \text{sum} &=& 85228 \end{array}$

3.

The probability is:

$\begin{array}{rcll} \frac{ 85228 } { 6^{10} } &=& \frac{ 85228 } { 60466176 } \\ &=& 0.00140951530 \end{array}$

Ein Würfel wird zweimal geworfen und das Produkt der beiden Augenzahlen notiert. ist es günstiger auf das Produkt 15 oder auf dass Produkt 12 zu wetten ?

$\begin{array}{|r|r|r|r|r|r|r|r|} \hline & \text{Wurf 2} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{Wurf 1} & \times & & & & & & \\ \hline 1 & & 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & & 2 & 4 & 6 & 8 & 10 & {\color{red}12}\\ 3 & & 3 & 6 & 9 & {\color{red}12} & {\color{green}15} & 18 \\ 4 & & 4 & 8 & {\color{red}12} & 16 & 20 & 24 \\ 5 & & 5 & 10 & {\color{green}15} & 20 & 25 & 30 \\ 6 & & 6 & {\color{red}12} & 18 & 24 & 30 & 36 \\ \hline \end{array}$

Es ist günstiger auf das Produkt 12 zu wetten, da es viermal vorkommen kann. Die 15 kann nur zweimal vorkommen.

12 = 2*6 = 3*4 = 4*3 = 6*2

15 = 3*5 = 5*3

log25+log2xlog50+(log2)^2=?

Vielen Dank asinus.

(Thank you asinus)

The volume of a cone w a height of 12 in. is 144pi cubic inches.

What is the length in inches, of the diameter of its base?

$h_{\text{cone}} = 12\ in.\\ V_{\text{cone}} = 144\cdot \pi\ in^3 \\ d_{\text{cone}} = 2\cdot r \\ r = radius$

$\begin{array}{rcll} \frac13 \cdot \pi \cdot r^2 \cdot h &=& V_{\text{cone}} \qquad &| \qquad h = 12 \quad V=144\cdot \pi\\ \frac13 \cdot \pi \cdot r^2 \cdot 12 &=& 144\cdot \pi \qquad &| \qquad :\pi \\ \frac13 \cdot r^2 \cdot 12 &=& 144 \qquad &| \qquad 144 = 12\cdot 12 \\ \frac13 \cdot r^2 \cdot 12 &=& 12\cdot 12 \qquad &| \qquad : 12 \\ \frac13 \cdot r^2 &=& 12 \qquad &| \qquad \cdot 3 \\ r^2 &=& 12 \cdot 3 \\ r^2 &=& 36 \qquad &| \qquad 36= 6\cdot 6 \\ r^2 &=& 6\cdot 6 \\ r^2 &=& 6^2 \qquad &| \qquad \sqrt{()} \\ r &=& 6 \\\\ d &=& 2\cdot r \\ d &=& 2\cdot 6 \\ \mathbf{ d }& \mathbf{=} & \mathbf{12\ in.} \end{array}$

The diameter of its base is 12 in.

log25+log2xlog50+(log2)^2=?

$\small{ \begin{array}{rcll} && \log(25)+ \log(2) \cdot \log(50)+[\log(2)]^2 \\ &=& \log(5^2)+ \log(2) \cdot \log(2\cdot 5^2)+ \log(2)\cdot \log(2) \\ &=& 2\cdot\log(5)+ \log(2) \cdot [~ \log(2) +\log(5^2) ~]+ \log(2)\cdot \log(2) \\ &=& 2\cdot\log(5)+ \log(2) \cdot \log(2) + \log(2)\cdot\log(5^2)+ \log(2)\cdot \log(2) \\ &=& 2\cdot\log(5)+ 2\cdot \log(2) \cdot \log(2) + \log(2)\cdot\log(5^2)\\ &=& 2\cdot\log(5)+ 2\cdot \log(2) \cdot \log(2) + 2\cdot \log(2)\cdot\log(5)\\ &=& 2\cdot\log(5)+ 2\cdot \log(2) \cdot [~ \log(2) + \log(5) ~]\\ &=& 2\cdot\log(5)+ 2\cdot \log(2) \cdot \log(2\cdot 5) \\ &=& 2\cdot\log(5)+ 2\cdot \log(2) \cdot \log(10) \qquad | \qquad \log_{10}(10) = 1\\ &=& 2\cdot\log(5)+ 2\cdot \log(2) \cdot 1\\ &=& 2\cdot\log(5)+ 2\cdot \log(2) \\ &=& 2\cdot [~ \log(5)+ \log(2) ~] \\ &=& 2\cdot \log(5\cdot 2) \\ &=& 2\cdot \log(10) \qquad | \qquad \log_{10}(10) = 1\\ &=& 2\cdot 1\\ &=& 2 \\ && \mathbf{ \log(25)+ \log(2) \cdot \log(50)+[\log(2)]^2 = 2 } \end{array} }$

Observe the pattern: 2,5,8,14,... if the pattern continues, what is the 41st number?

$\begin{array}{r|r} \hline & \text{Primes of the form 4n-1 } & \\ & p & \frac{3\cdot p - 1}{4} \\ \hline 1 & 3 & a_1 = \frac{3\cdot 3 - 1}{4} =2 \\ 2 & 7 & a_2 = \frac{3\cdot 7 - 1}{4} =5 \\ 3 & 11 & a_3 = \frac{3\cdot 11 - 1}{4} = 8 \\ 4 & 19 & a_4 = \frac{3\cdot 19 - 1}{4} = 14 \\ 5 & 23 & a_5 = \frac{3\cdot 23 - 1}{4} = 17 \\ 6 & 31 & 23 \\ 7 & 43 & 32 \\ 8 & 47 & 35 \\ 9 & 59 & 44 \\ 10 & 67 & 50 \\ 11 & 71 & 53\\ 12 & 79 & 59 \\ 13 & 83 & 62 \\ 14 & 103 & 77 \\ 15 & 107 & 80 \\ 16 & 127 & 95 \\ 17 & 131 & 98\\ 18 & 139 & 104 \\ 19 & 151 & 113\\ 20 & 163 & 122 \\ 21 & 167 & 125 \\ 22 & 179 & 134 \\ 23 & 191 & 143 \\ 24 & 199 & 149 \\ 25 & 211 & 158 \\ 26 & 223 & 167 \\ 27 & 227 & 170 \\ 28 & 239 & 179\\ 29 & 251 & 188\\ 30 & 263 & 197 \\ 31 & 271 & 203 \\ 32 & 283 & 212 \\ 33 & 307 & 230 \\ 34 & 311 & 233 \\ 35 & 331 & 248 \\ 36 & 347 & 260 \\ 37 & 359 & 269 \\ 38 & 367 & 275 \\ 39 & 379 & 284 \\ 40 & 383 & 287 \\ 41 & 419 & a_{41} = \frac{3\cdot 419 - 1}{4} = 314 \\ 42 & 431 & 323 \\ 43 & 439 & 329 \\ 44 & 443 & 332 \\ 45 & 463 & 347 \\ \cdots & \cdots & \cdots \\ \hline \end{array}$

Evaluate without a calculator:

$[~12\binom{3}{3} + 11\binom{4}{3} + 10\binom{5}{3} + \cdots + 2\binom{13}{3} +\binom{14}{3} ~]$

Identity:

$\text{For }~ n,r \in N^+, r \le n,\\ \dbinom{n+1}{r+1} = \sum \limits_{j=r}^{n} \dbinom{j}{r}$

if i have sides with each having a length of 13, 14, and 15. how big will each angle be?

Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is
$A = \sqrt{s(s-a)(s-b)(s-c)},$
where s is the semiperimeter of the triangle; that is,
$s=\frac{a+b+c}{2}$

$\begin{array}{rcl} s &=& \frac{a+b+c}{2} \\ &=& \frac{13+14+15}{2} \\ &=& \frac{42}{2} \\ \mathbf{s} & \mathbf{=}& \mathbf{21} \\\\ A &=& \sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-c)} \\ &=& \sqrt{21\cdot (21-13)\cdot (21-14)\cdot (21-15)} \\ &=& \sqrt{21\cdot 8 \cdot 7 \cdot 6} \\ &=& \sqrt{7056} \\ \mathbf{A} & \mathbf{=}& \mathbf{84} \end{array}$

$\begin{array}{rcl} 2\cdot A &=& b\cdot c\cdot \sin \alpha \\ \sin \alpha &=& \frac{2\cdot A}{ b\cdot c } \\ \sin \alpha &=& \frac{2\cdot 84}{ 14\cdot 15 } \\ \alpha &=& \arcsin{ (\frac{168}{ 14\cdot 15 } ) } \\ \alpha &=& \arcsin{ (0.8) } \\ \mathbf{\alpha }&\mathbf{=}& \mathbf{53.1301023542^{\circ}} \end{array}$

$\begin{array}{rcl} 2\cdot A &=& c\cdot a\cdot \sin \beta \\ \sin \beta &=& \frac{2\cdot A}{ c\cdot a } \\ \sin \beta &=& \frac{2\cdot 84}{ 15\cdot 13 } \\ \beta &=& \arcsin{ (\frac{168}{ 15\cdot 13 } ) }\\ \beta &=& \arcsin{ (0.86153846154) }\\ \mathbf{\beta} &\mathbf{=}& \mathbf{59.4897625939^{\circ}} \end{array}$

$\begin{array}{rcl} 2\cdot A &=& a\cdot b\cdot \sin \gamma \\ \sin \gamma &=& \frac{2\cdot A}{ a\cdot b} \\ \sin \gamma &=& \frac{2\cdot 84}{ 13\cdot 14 } \\ \gamma &=& \arcsin{ (\frac{168}{ 13\cdot 14 }) }\\ \gamma &=& \arcsin{ (0.92307692308) }\\ \mathbf{\gamma }&\mathbf{=}& \mathbf{67.3801350520^{\circ}} \end{array}$

Bitte um Hilfe: Wie kann man folgende Angabe mit einer Gleichung lösen?

In einem Geschäft gibt es blaue, grüne, weiße und schwarze Pullover. Es sind halb so viele blaue wie grüne Pullover. Es sind halb so viele grüne wie weiße Pullover. Von den schwarzen Pullovern gibt es 18 Stück, das sind halb so viele wie von den grünen und weißen Pullovern zusammen. Wie viele Pullover gibt es in diesem Geschäft?

Ich habe zwar die Lösung von der Anzahl der Pullovers. Ich weiß aber leider nicht, wie ich mit einer oder zwei Gleichungen diese Aufgabe lösen kann. Bitte um dringende Hilfe! Vielen Dank!!!

b = blau

g = grün

w = weiß

s = schwarz

$\begin{array}{lrcl} (1) & g &=& 2b \\ \hline (2) & w &=& 2g \\ & w &=& 2\cdot 2b \\ & w &=& 4b\\ \hline (3) & s &=& 18 \\ \hline (4) & g+w &=& 2s \\ & g+w &=& 2\cdot 18 \\ & g+w &=& 36 \\ \hline (5) & g + w &=& 2b+4b\\ & g + w &=& 6b\\ \hline (4)=(5) & g+w &=& 36 \\ & g+w &=& 6b \\ & 6b &=& 36 \\ & b &=& \frac{36}{6} \\ & \mathbf{b} & \mathbf{=} & \mathbf{6} \\ \hline & g &=& 2b\\ & g &=& 2\cdot 6\\ & \mathbf{g} & \mathbf{=} & \mathbf{12} \\ \hline & w &=& 2g\\ & w &=& 2\cdot 12\\ & \mathbf{w} & \mathbf{=} & \mathbf{24} \\ \hline & b+g+w+s &=& 6+12+24+18\\ & &=& 60 \\ & \mathbf{b+g+w+s} & \mathbf{=} & \mathbf{60} \end{array}$

Die Anzahl der Pullover sind 60.

Es gibt 6 blaue Pullover.

Es gibt 12 grüne Pullover.

Es gibt 24 weiße Pullover.

Es gibt 18 schwarze Pullover.

Find the area of the shaded region below.

$\begin{array}{rcll} f(y) &=& e^y \\ g(y) &=& y^2-2 \\\\ A &=& \int \limits_{y=-1}^{y=1} { [f(y)-g(y)]\ dy}\\\\ A &=& \int \limits_{y=-1}^{y=1} { [ e^y-(y^2-2)]\ dy}\\\\ A &=& \int \limits_{y=-1}^{y=1} { ( e^y-y^2+2 )\ dy}\\\\ A &=& [~ e^y-\frac{y^3}{3}+2y ~]_{y=-1}^{y=1} \\ A &=& e^1-\frac{1^3}{3}+2\cdot 1 -(e^{-1}-\frac{(-1)^3}{3}+2\cdot(-1) ) \\ A &=& e -\frac13 +2 - ( \frac{1}{e}+\frac13 - 2) \\ A &=& e -\frac13 +2 - \frac{1}{e} - \frac13 + 2 \\ A &=& e - \frac{1}{e} -\frac23 + 4 \\ A &=& e - \frac{1}{e} + \frac{10}{3} \\ A &=& 2.71828182846 - 0.36787944117 + 3.33333333333 \\ \mathbf{A} & \mathbf{=} & \mathbf{ 5.68373572062 } \end{array}$