Probability/ Combination Question | Help!

**mrsenim** · April 5th 2010, 01:43 AM

Hello,

I am new to probability.

Please have a look at the following question.

We have n = 10 persons and we wish to divide them at random into 3 groups consisting of 5, 3, and 2 persons respectively. In how many ways is this possible?

I have tried to solve it in following way.

Total number of possible ways =

$ 10C5 = \left( 10! \over 5!(10-5)! \right) = \left( 10 \times 9 \times 8 \times 7 \times 6 \over 5 \times 4 \times 3 \times 2 \right) = \left( 30240 \over 120 \right) = 252 $

Next!

$ 5C3 = \left( 5! \over 3!(5-3)! \right) = \left( 5 \times 4 \over 2 \right) = \left( 20 \over 2 \right) = 10 $

Next!

$ 2C2 = \left( 2! \over 2!(2-2)! \right) = 1 $

$ S = 10C5 \times 5C3 \times 2C2 = 252 \times 10 \times 1 = 2520 $

I think the above solution is correct but I really don't know how it happens.

I mean to say that I am confused in

1 - Why are we multiplying 10C5 with 5C3 and 2C2 ? Which rule or formula implies this multiplication of combinations?

2 - When and in which situations do I need to multiply combinations?

Any help would be highly appreciated.

**Archie Meade** · April 5th 2010, 02:42 AM

Originally Posted by mrsenim

Hello,

I am new to probability.

Please have a look at the following question.

We have n = 10 persons and we wish to divide them at random into 3 groups consisting of 5, 3, and 2 persons respectively. In how many ways is this possible?

I have tried to solve it in following way.

Total number of possible ways =

$ 10C5 = \left( 10! \over 5!(10-5)! \right) = \left( 10 \times 9 \times 8 \times 7 \times 6 \over 5 \times 4 \times 3 \times 2 \right) = \left( 30240 \over 120 \right) = 252 $

Next!

$ 5C3 = \left( 5! \over 3!(5-3)! \right) = \left( 5 \times 4 \over 2 \right) = \left( 20 \over 2 \right) = 10 $

Next!

$ 2C2 = \left( 2! \over 2!(2-2)! \right) = 1 $

$ S = 10C5 \times 5C3 \times 2C2 = 252 \times 10 \times 1 = 2520 $

I think the above solution is correct but I really don't know how it happens.

I mean to say that I am confused in

1 - Why are we multiplying 10C5 with 5C3 and 2C2 ? Which rule or formula implies this multiplication of combinations?

2 - When and in which situations do I need to multiply combinations?

Any help would be highly appreciated.

Hi mrsenim,

$10c_5=\binom{10}{5}$

is the number of ways of choosing a group of 5 from 10 available.

Taking one of these groups of 5, there will be 5 remaining.

$5c_3=\binom{5}{3}$

is the number of ways to choose 3 from the remaining 5
(the 2 left over will always be unique to this 3).

Hence any selection of 5 from 10 can be matched with any one of these $5c_3$ groups of 3 and 2.

Since there are $10c_5$ groups of 5 in total,
we count the number of group matches by multiplying the number
of groups of 5 by the number of groups of 3.

**mrsenim** · April 5th 2010, 09:19 PM

Originally Posted by Archie Meade

Hi mrsenim,

$10c_5=\binom{10}{5}$

is the number of ways of choosing a group of 5 from 10 available.

Taking one of these groups of 5, there will be 5 remaining.

$5c_3=\binom{5}{3}$

is the number of ways to choose 3 from the remaining 5
(the 2 left over will always be unique to this 3).

Hence any selection of 5 from 10 can be matched with any one of these $5c_3$ groups of 3 and 2.

Since there are $10c_5$ groups of 5 in total,
we count the number of group matches by multiplying the number
of groups of 5 by the number of groups of 3.

Thank You, Archie Meade for replying to my question. I think the reply is quite understandable.

But I am still confused about multiplying the two combination that is
$10c_5 \times 5c_3$ to calculate the total number of group matches.

Could you please provide some link or documentation which clarifies this concept in detail?

Thanks Again.

Yours.

**Failure** · April 5th 2010, 09:25 PM

Originally Posted by mrsenim

But I am still confused about multiplying the two combination that is
$10c_5 \times 5c_3$ to calculate the total number of group matches.

Could you please provide some link or documentation which clarifies this concept in detail?

What about Rule of product - Wikipedia, the free encyclopedia

**Archie Meade** · April 6th 2010, 02:43 AM

Hi mrsenim,

I'm not sure about a good link,
but here's a way to view it.

It's easier to understand with small groups.

Let's say we pick groups of 3 and 2 from a total of 6.

There are $6c_2=15$ ways to pick 2.

Label these groups A, B, C, D, E, F, G, H, I, J, K, L, M, N, O.

Now take group A.
Having chosen the 2, there are 4 remaining from which to choose another different group of 2.

The second pair can be chosen in $4c_2=6$ ways.

Label these groups U, V, W, X, Y, Z.

Group A may be paired with any of the groups U, V, W, X, Y, Z.
That's 6 pairings containing group A.

If we take the group B, we can form another 6 groups from the 4 remaining
after the 2 people in group B have been chosen.

Same for the other groups C to O.
Hence, there are 6 pairings 15 times.

Therefore there are (15)(6) pairings.

This is $\binom{6}{2}\binom{4}{2}$

**mrsenim** · April 6th 2010, 09:14 PM

Originally Posted by Archie Meade

Hi mrsenim,

Same for the other groups C to O.
Hence, there are 6 pairings 15 times.

Therefore there are (15)(6) pairings.

This is $\binom{6}{2}\binom{4}{2}$

ohh! ok. Thank you Archie Meade, it's clear now.

Originally Posted by Failure

What about Rule of product - Wikipedia, the free encyclopedia

Thank you this link is helpful.

**matheagle** · April 6th 2010, 11:27 PM

The best way to see
10 persons and we wish to divide them at random into 3 groups consisting of 5, 3, and 2
is to think of 10 slots, there are 10! ways of placing 10 items in these 10 positions.
Then break those 10 into three groups, the first 5, the next 3 and the last 2.
You end up with "double counting" of 5! in the first 5 spots since any rearrangement of those 5 is still the same.
Likewise for the next two groups.
Hence the answer is

${10\choose 5,3,2}={10!\over 5! 3! 2!}$

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