In how many different ways can a group of 9 people be divide

To divide 9 persons into 3 groups, when the ordering of groups is not important can be done in \(\frac{1}{3!} * \frac{9!}{(3!)^3}\) ways.

Answer is (A) or 280

the number of ways to choose 9 people in 3 groups each having 3 people is

9C3 * 6C3 * 3C3 = 280

another way is = (3*3)!/((3!)^3)*3! = 280

Hi Bunuel
Can u pls explain why u r dividing by 3!
we are not ordering here we are only choosing.

How can we compute the # of ways in which 9 objects are divided into groups of 4,3 and 2 ? Can you please help ?

Here's what I think:

let 9 objects be AAAABBBCC
Therefore, combinations = 9!/(4!*3!*2!) Correct ? [Order is important]

If order is not important,

since we have three groups,

# of combinations = 9!/[(4!*3!*2!) * (3!)]

Correct ?

Actually, in this case the groups are distinct - a group of 4 people, another of 3 people and another of 2 people. A case in which Mr A is in the four person group is different from the one in which he is in 3 person group. So you will not divide by 3! at the end in second case.

Thanks Karishma. So, are you saying that the order will not matter ? Essentially, # of combinations = 9!/(4!*3!*2!) irrespective of order?/

Thanks
Voodoo

Yes, the groups are distinct so no of combinations is 9!/(4!*3!*2!)

Hi Geturdream,
How comes A = 9C3 * 6C3 * 3C3 = 280 ?
9C3 = 9!/ (6!*3!) = 9*8*7 / (3*2) = 84
6C3 = 6! / (3!*3!) = 6*5*4 / (3*2) = 20
3C3 = 3!/ (3!*0!) = 1
=> A = 84 * 20 = 1680 ?

For those who don't understand why we need to divide by 3!

Let say we have 3 people in blue pants, 3 in red pants, and 3 in green pants.

\(C^9_3*C^6_3*C^3_3\) have all following 6 combinations:

[P P P] [P P P] [P P P]
[P P P] [P P P] [P P P]
[P P P] [P P P] [P P P]
[P P P] [P P P] [P P P]
[P P P] [P P P] [P P P]
[P P P] [P P P] [P P P]

But in fact all those combinations represent only 1 outcome and that is why we need to exclude order (divide by 3!).


Here is a quick review of fundamentals: math-combinatorics-87345.html

I used a way found in another topic:

How many ways can 1 person be put with the other 8 in groups of 3? 28
How many ways can following person be put with the remaining 5 in groups of 3? (subract first group total) : 10
How many ways can the final 3 be placed into a group of 3? (subtract last 3) : 1
28 * 10 * 1 = 280
answer is A

1680 needs to be divided by 3! because the order of the 3 groups do not matter for this problem. The order of the groups, i.e. G1G2G3 vs G2G3G1 (there are 4 more possible combinations) do not matter.

\(\frac{9C3 * 6C3 * 3C3}{3!} = \frac{84*20*1}{3!} = \frac{1680}{3!} = 280\)

I don't think that's right. For example, let's find out in how many ways 4 different items can be split into 2 groups of 2, when the order is NOT important, using the 2nd formula.

\(\frac{(mn)!}{(n!)^m*m!}\) = \(\frac{(4)!}{(2!)^2*2!}\) = 3

This result can be disproved very easily graphically:

1. Draw a square
2. Draw an x through the square drawing 2 lines, corner to corner
3. Represent the 4 objects on each corner of the square
4. Number the lines that link all the square's corners together

There are 6 lines which represent all 6 possible groups of 2 objects; every object is connected by a line to all other objects a single time, forming 6 possible groups of 2.

The answer is 6.

Another solution is 4C2 * 2C2 = 6

nCr = \(\frac{(n!)}{(r!(n-r)!)}\) = \(\frac{(4!)}{(r!(4-2)!)}\) * \(\frac{(2!)}{(r!(2-2)!)}\) = 6

The groups have not been ordered yet and there is no need to divide by anything else; we proved graphically that 6 is the answer so dividing by 2! or anything else would lead to an incorrect result.
If the order was important, there would be \(6!/(6-2)! = 30\) ways to organize 4 objects into 2 groups of 2.[/b]

There would be 2 positions to fill with 6 possible groups. 6 possible groups in the 1st position and 5 possible groups in the 2nd position.

nPr= \(\frac{n!}{(n-r)!}\) = \(\frac{6!}{(6-2)!}\) = 6 * 5 = 30

The 1st formula actually represents the number of ways in which the order is NOT important

\(\frac{(mn)!}{(n!)^m}\) = \(\frac{(4)!}{(2!)^2}\) = 6

THE ORIGINAL QUESTION:

The order is not important in this question.

nCr = \(\frac{(n!)}{(r!(n-r)!)}\) = \(\frac{(9!)}{(3!(9-3)!)}\) * \(\frac{(6!)}{(3!(6-3)!)}\) * \(\frac{(3!)}{(3!(3-3)!)}\) = 1680


or, alternatively:


\(\frac{(mn)!}{(n!)^m}\) = \(\frac{(9)!}{(3!)^3}\) = 1680

If the order mattered:

nPr= \(\frac{n!}{(n-r)!}\) = \(\frac{1680!}{(1680-3)!}\) = 1680 * 1679 * 1678 = 4733168160

No you are not right.

There are two formulas. The first one for which the order of the groups IS important and the second one for which the order of the groups is NOT important.

Case 1 - the order of the groups IS important: \(\frac{(mn)!}{(n!)^m}\)

2*2 = 4 different items can be divided equally into 2 groups and the order of the groups IS important \(\frac{4!}{(2!)^2}=6\).

GROUP 1 - GROUP 2
{AB} - {CD}
{CD} - {AB}

{AC} - {BD}
{BD} - {AC}

{AD} - {BC}
{BC} - {AD}

Case 2 - the order of the groups is NOT important: \(\frac{(mn)!}{(n!)^m * m!}\)

2*2 = 4 different items can be divided equally into 2 groups and the order of the groups is NOT important \(\frac{4!}{(2!)^2*2}=3\).

{AB} - {CD}

{AC} - {BD}

{AD} - {BC}

P.S. The correct answer is A, not C. Please refer to several different solutions presented above.

Can we apply the same logic to the below question.
A menu consisting of 2 types of breakfasts, 3 types of lunch and 4 types of dinner has to be selected from 10 types of breakfast, 11 types of lunch and 12 types of dinner.
Will the solution be 10c*11c3*12c4/3! ?

Can we apply the same logic to the below question.
A menu consisting of 2 types of breakfasts, 3 types of lunch and 4 types of dinner has to be selected from 10 types of breakfast, 11 types of lunch and 12 types of dinner.
Will the solution be 10c*11c3*12c4/3! ?

We can start with this combination problem, like most of them, by looking at one part of it. How many ways are there of forming the first group or pile? That’s choosing 3 people from 9. Applying the combination formula, with n = 9 and k = 3,

\(9C3\) = 84

To form the next group, we compute

\(6C3\) = 20

There are three people from the original group remaining, and only one group of three left, so there is only one way to finish. That means there are 84 × 20 × 1 = 1,680 ways to create the three groups.

However, that number, 1,680, involves double-counting some groups: it counts ABC DEF GHI and DEF ABC GHI as different ways to do the grouping, but for our purposes those ways are the same, since the three groups that we are placing people into are indistinguishable. For each way, the duplication will be the number of ways of shuffling the three groups around, which is 3! = 6. The number of unique ways to form the groups is therefore . The Correct Answer is (A).

The first group of 3 can be chosen in 9C3 = (9 x 8 x 7)/(3 x 2) = 3 x 4 x 7 = 84 ways.

The second group of 3 can be chosen in 6C3 = (6 x 5 x 4)/(3 x 2) = 5 x 4 = 20 ways.

The third group of 3 can be chosen in 3C3 = 1 way.

Therefore, the 3 groups can be chosen 84 x 20 x 1 = 1680 ways. However, since the order of the 3 groups doesn’t matter, we have to divide 1680 by 3!. Hence, the number of ways 9 people can be divided into 3 groups is 1680/3! = 1680/6 = 280.

Answer: A