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Neko [114]
2 years ago
15

Dave has 10 poker chips, 5 of which are red and the other 5 of which are white. Dave likes to stack his chips and flip them over

as he plays. How many different 10-chip stacks can Dave make if two stacks are not considered distinct if one can be flipped to appear identical to the other?
Mathematics
2 answers:
sergey [27]2 years ago
7 0

Answer:

\frac{10!}{5!*5!}  -  \frac{5!}{3!*2!}  

Step-by-step explanation:

Given:

10 poker chips:

  • 5 red
  • 5 white

Let the stack has positions (1,2,3,4,5,6,7,8,9,10)

So we can find all possible outcomes of the stack is: \frac{10!}{5!*5!} =

and the possibility of a stack to be identical while flipping is: \frac{5!}{3!*2!}

So the different 10-chip stacks can Dave make if two stacks are not considered distinct:

\frac{10!}{5!*5!}  - \frac{5!}{3!*2!}  

max2010maxim [7]2 years ago
7 0

Answer:

10!/(5!× 5!) - 5!/(3! × 2!)

Step-by-step explanation:

From the question,Dave has 10 poker chips, 5 of which are red and the other 5 of which are white. Dave likes to stack his chips and flip them over as he plays. How many different 10-chip stacks can Dave make if two stacks are not considered distinct if one can be flipped to appear identical to the other?

-----Here we have 10 chips

-----5 are red and the rest is white

We are now asked to find out how many different 10-chip stacks can Dave make if two stacks are not considered distinct if one can be flipped to appear identical to the other.

Let the stack has positions (1,2,3,4,5,6,7,8,9,10)

All possibility of stacks is

10!/(5! × 5!)

and the possibility of a stack to be identical while flipping is

5!/(3! × 2!)

No of different stacks is

10!/(5! × 5!) - 5!/(3! × 2!)

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alex41 [277]
Procedure:

1) calculate the number of diferent teams of four members that can be formed (with the ten persons)
2) calculate the number of teams tha meet the specification (two girls and two boys)
3) Divide the positive events by the total number of events: this is the result of 2) by the result in 1)

Solution

1) the number of teams of four members that can be formed are:

10*9*8*7 / (4*3*2*1) = 210

2) Number of different teams with 2 boys and 2 girls = ways of chosing 2 boys * ways of chosing 2 girls

Ways of chosing 2 boys = 6*5/2 = 15
Ways of chosing 2 girls = 4*3/2 = 6

Number of different teams with 2 boys and 2 girls = 15 * 6 = 90

3) probability of choosing one of the 90 teams formed by 2 boys and 2 girls:

90/210 = 3/7
 
7 0
3 years ago
Simplify the product using the distributive property. (-4h +2)(3h +7)
Alekssandra [29.7K]

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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scZoUnD [109]
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The length of time required for the periodic maintenance of an automobile or another machine usually has a mound-shaped probabil
maks197457 [2]

Solution :

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Standard deviation = 0.7 hours

Maximum average service time = 1.6 hours for one automobile

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