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Nine dogs are split into 3 groups to pull one of three sled
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gmatter2012
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Nine dogs are split into 3 groups to pull one of three sleds in a race. How many different assignments of dogs to sleds are possible?
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adthedaddy
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Can you please share the OA and the source ?
I shall solve it this way -
3 groups of 3 dogs are to be formed from 9 dogs.
We can create first group of 3 dogs from 9 dogs in 9C3 ways.
Of the remaining 6, we can create the 2nd group of 3 dogs in 6C3 ways.
And finally, the remaining 3 dogs' group can be formed in 1 way (3C3).
Therefore, total assignments = 9C3*6C3*3C3 = 84*20*1 = 1680
I shall solve it this way -
3 groups of 3 dogs are to be formed from 9 dogs.
We can create first group of 3 dogs from 9 dogs in 9C3 ways.
Of the remaining 6, we can create the 2nd group of 3 dogs in 6C3 ways.
And finally, the remaining 3 dogs' group can be formed in 1 way (3C3).
Therefore, total assignments = 9C3*6C3*3C3 = 84*20*1 = 1680
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sanrisenew
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9c3 = no. of ways to select 3 dogs out of 9
3c1 = no. of ways of selecting one sledge out of 3
Therefore total no. of ways to apply a team of three dogs in first sledge
= 9c3* 3c1
for second set of dogs
= 6c3*2c1
for third set
= 3c3*1c1
Therefore total ways = 9c3*3c1+6c3*2c1+ 3c3*1c1
3c1 = no. of ways of selecting one sledge out of 3
Therefore total no. of ways to apply a team of three dogs in first sledge
= 9c3* 3c1
for second set of dogs
= 6c3*2c1
for third set
= 3c3*1c1
Therefore total ways = 9c3*3c1+6c3*2c1+ 3c3*1c1
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gmatter2012
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adthedaddy wrote:Can you please share the OA and the source ?
I shall solve it this way -
3 groups of 3 dogs are to be formed from 9 dogs.
We can create first group of 3 dogs from 9 dogs in 9C3 ways.
Of the remaining 6, we can create the 2nd group of 3 dogs in 6C3 ways.
And finally, the remaining 3 dogs' group can be formed in 1 way (3C3).
Therefore, total assignments = 9C3*6C3*3C3 = 84*20*1 = 1680
That seems correct, was not sure whether we have to multiply 3! at the end.
In this case I believe you have taken order to be important.
But what if the order was not important how do you think the answer could change, like in the question below ?
In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?
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LalaB
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well, yes, in this case order is important.
Remember-
if u want to get m groups, each containing n objects, and order matters, then use the formula - (mn)!/(n!)^m
so, in this case we have (3*3)!/(3!)^3
if u want to get m groups, each containing n objects, and order doesnt matter, then use the formula - (mn)!/(n!)^m * m!
it could be (3*3)!/(3!)^3*3!
that is all:)
Remember-
if u want to get m groups, each containing n objects, and order matters, then use the formula - (mn)!/(n!)^m
so, in this case we have (3*3)!/(3!)^3
if u want to get m groups, each containing n objects, and order doesnt matter, then use the formula - (mn)!/(n!)^m * m!
it could be (3*3)!/(3!)^3*3!
that is all:)
Happy are those who dream dreams and are ready to pay the price to make them come true.(c)
In order to succeed, your desire for success should be greater than your fear of failure.(c)
In order to succeed, your desire for success should be greater than your fear of failure.(c)
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gmatter2012
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LalaB
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u r welcome. glad to be helpful, and good luck with ur studygmatter2012 wrote:Thank you LalaB, these formula's, Good to know.
Happy are those who dream dreams and are ready to pay the price to make them come true.(c)
In order to succeed, your desire for success should be greater than your fear of failure.(c)
In order to succeed, your desire for success should be greater than your fear of failure.(c)
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gmatter2012
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can you please keep track of my questions , Am suffering a lot lately due to combinations, would really appreciate your help ( have also asked @kanwar for his assistance).LalaB wrote:u r welcome. glad to be helpful, and good luck with ur studygmatter2012 wrote:Thank you LalaB, these formula's, Good to know.
I have found quite a few combinotrics questions whose solutions I am not sure about, Hope I will be able to find help in this community. Again looking forward to your assistance.Thank you
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LalaB
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gmatter2012 wrote:sure. we are all here to find help and at the same time to be helpfulLalaB wrote:
can you please keep track of my questions , Am suffering a lot lately due to combinations, would really appreciate your help ( have also asked @kanwar for his assistance).
I have found quite a few combinotrics questions whose solutions I am not sure about, Hope I will be able to find help in this community. Again looking forward to your assistance.Thank you
Happy are those who dream dreams and are ready to pay the price to make them come true.(c)
In order to succeed, your desire for success should be greater than your fear of failure.(c)
In order to succeed, your desire for success should be greater than your fear of failure.(c)
