Nine dogs are split into 3 groups to pull one of three sled

Question

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?

adthedaddy · Answer

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

sanrisenew · Answer

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

gmatter2012 · Answer

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?

LalaB · Answer

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:)

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gmatter2012 · Answer

Thank you LalaB, these formula's, Good to know.

LalaB · Answer

gmatter2012 wrote:Thank you LalaB, these formula's, Good to know. u r welcome. glad to be helpful, and good luck with ur study

gmatter2012 · Answer

LalaB wrote:gmatter2012 wrote:Thank you LalaB, these formula's, Good to know. u r welcome. glad to be helpful, and good luck with ur study 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

LalaB · Answer

gmatter2012 wrote:LalaB 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 sure. we are all here to find help and at the same time to be helpful

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