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egoroff_w [7]
3 years ago
8

Timothy is re-arranging his marble collection.he has five identical blue marbles,five identical green marbles and three identica

l black marbles he can fit exactly five marbles into a case and must have at least one of each.How many different ways can he arrange the case in?
Mathematics
2 answers:
romanna [79]3 years ago
7 0
Based on the scenario, the case must be filled like this :

blue/green/black  x   blue/ green/black    x blue/green/black   x  random   x random

Different ways he can arrange it : 

3 x 5!/(2!x2!)       +            3x    <span>5!/3!
</span>
90           +        60

= 150 ways

hope this helps







Ksenya-84 [330]3 years ago
6 0

Answer:

720

Step-by-step explanation:

Knowing that he must place one of each type, the total number of marbles must be one of theses 6 cases:

3 blues, 1 green, 1 black

2 blues, 2 green, 1 black

2 blues, 1 green, 2 black

1 blues, 2 green, 2 black

1 blues, 1 green, 3 black

1 blues, 3 green, 1 black

At the same time, each case can be ordered in many ways, for example in the first option:

blue, blue, blue, green, black

blue, blue, blue, black, green

blue, blue, black, blue, green

et cetera

The total number of permutations for each case is 5! = 120

Taking into account the 6 cases, 6 x 120 = 720

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Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E
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Answer:

(A) y=ke^{2t} with k\in\mathbb{R}.

(B) y=ke^{2t}/2-1/2 with k\in\mathbb{R}

(C) y=k_1e^{2t}+k_2e^{-2t} with k_1,k_2\in\mathbb{R}

(D) y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5 with k_1,k_2\in\mathbb{R},

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then 0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow  \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}. With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form e^{2t} with t real.

(B) Proceeding and the previous item, we obtain 1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow  \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2. Which is not a vector space with the usual operations (this is because -1/2), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set y=e^{mt} \Rightarrow y''=m^2e^{mt} and we obtain the characteristic equation 0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2 and then the general solution is y=k_1e^{2t}+k_2e^{-2t} with k_1,k_2\in\mathbb{R}, and as in the first items the set of solutions form a vector space.

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