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Neko [114]
3 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]3 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]3 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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