Combination Question Watch

bigmansouf
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#1
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Q:

In how many ways can 8 white and 4 black draughtsmen can be arranged in a pile?


I am having difficulty with this. I do not know where to start

8 w 4 b

in a pile so does it mean one for after the other like layering one brick onto of another one
my problem is it does not say how many spaces in a pile so i am confused

please can i have some direction
thank you
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mqb2766
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(Original post by bigmansouf)
Q:

In how many ways can 8 white and 4 black draughtsmen can be arranged in a pile?


I am having difficulty with this. I do not know where to start

8 w 4 b

in a pile so does it mean one for after the other like layering one brick onto of another one
my problem is it does not say how many spaces in a pile so i am confused

please can i have some direction
thank you
Think of just one colour and how many positions can be allocated to each counter of that colour.

Edit: I think they mean a stack of12 counters with different orderings of the white and black counters. No spaces.
Last edited by mqb2766; 4 weeks ago
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RDKGames
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(Original post by bigmansouf)
Q:

In how many ways can 8 white and 4 black draughtsmen can be arranged in a pile?


I am having difficulty with this. I do not know where to start

8 w 4 b

in a pile so does it mean one for after the other like layering one brick onto of another one
my problem is it does not say how many spaces in a pile so i am confused

please can i have some direction
thank you
There are 12 in total. So there are 12! different arrangements.

But considering 8 whites are indistinguishable, and 4 blacks are indistinguishable, you need to adjust your answer accordingly.
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bigmansouf
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(Original post by RDKGames)
There are 12 in total. So there are 12! different arrangements.

But considering 8 whites are indistinguishable, and 4 blacks are indistinguishable, you need to adjust your answer accordingly.
(Original post by mqb2766)
Think of just one colour and how many positions can be allocated to each counter of that colour.

Edit: I think they mean a stack of12 counters with different orderings of the white and black counters. No spaces.
thank you for the help but the problem I am having is using combinations to answer it
but using permutations the answer is  \frac{12!}{8!4!}
but I dont see how to use combinations to answer this question

there are 12 spaces _ _ _ _ _ _ _ _ _ _ _ _
Now if I put the blacks this way
B _ _ B _ _ B _ _ B _ _
there are 8 spaces and if i choose 8 of the whites to fill in the spaces thus; 8C8 = 1

is there any way to do this using combinations theory only?
Last edited by bigmansouf; 4 weeks ago
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the bear
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(Original post by bigmansouf)
thank you for the help but the problem I am having is using combinations to answer it
but using permutations the answer is  \fac{12!}/({8!4!})
but I dont see how to use combinations to answer this question

there are 12 spaces _ _ _ _ _ _ _ _ _ _ _ _
Now if I put the blacks this way
B _ _ B _ _ B _ _ B _ _
there are 8 spaces and if i choose 8 of the whites to fill in the spaces thus; 8C8 = 1

is there any way to do this using combinations theory only?
looks like you overlooked the division
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mqb2766
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Think of the combinations of the group of positions of the black counters, for instance.
You get the other colour for free.
(Original post by bigmansouf)
thank you for the help but the problem I am having is using combinations to answer it
but using permutations the answer is  \fac{12!}{8!4!}
but I dont see how to use combinations to answer this question

there are 12 spaces _ _ _ _ _ _ _ _ _ _ _ _
Now if I put the blacks this way
B _ _ B _ _ B _ _ B _ _
there are 8 spaces and if i choose 8 of the whites to fill in the spaces thus; 8C8 = 1

is there any way to do this using combinations theory only?
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bigmansouf
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#7
Report Thread starter 4 weeks ago
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thanks i was able to solve it 12C8 = 495
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