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In how many different ways can a group of 9 people be divide 26 Sep 2010, 09:47
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In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?

A. 280
B. 1,260
C. 1,680
D. 2,520
E. 3,360
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In how many different ways can a group of 9 people be divide 26 Sep 2010, 09:56
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hemanthp
In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?

280
1,260
1,680
2,520
3,360
Show more

GENERAL RULE:
1. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\)

2. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is NOT important is \(\frac{(mn)!}{(n!)^m*m!}\).

BACK TO THE ORIGINAL QUESTION:
In original question I think the order is NOT important, as we won't have group #1, #2 and #3. So we should use second formula, \(mn=9\), \(m=3\) groups \(n=3\) objects (people):
\(\frac{(mn)!}{(n!)^m*m!}=\frac{9!}{(3!)^3*3!}=280\).

This can be done in another way as well: \(\frac{9C3*6C3*3C3}{3!}=280\), we are dividing by \(3!\) as there are 3 groups and order doesn't matter.

Answer: A.
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In how many different ways can a group of 9 people be divide 15 Jun 2011, 20:06
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toughmat
Hi Bunuel
Can u pls explain why u r dividing by 3!
we are not ordering here we are only choosing.
Show more

We divide by 3! because of exactly what you said: "we are not ordering here we are only choosing."
When you say, "9C3 * 6C3 * 3C3," what you are doing is that you are choosing 3 people of 9 for group 1, 3 people out of the leftover 6 people for group 2 and the rest of the three people for group 3. You have inadvertently marked the 3 groups as distinct. But if we want to just divide them in 3 groups without any distinction of group 1, 2 or 3, we need to divide this by 3!.
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In how many different ways can a group of 9 people be divide 02 Oct 2010, 00:41
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hemanthp
In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?

280
1,260
1,680
2,520
3,360
Show more

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
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In how many different ways can a group of 9 people be divide 03 Oct 2010, 04:25
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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
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In how many different ways can a group of 9 people be divide 15 Jun 2011, 06:34
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Hi Bunuel
Can u pls explain why u r dividing by 3!
we are not ordering here we are only choosing.
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In how many different ways can a group of 9 people be divide 18 Jun 2011, 22:14
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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 ? :)
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In how many different ways can a group of 9 people be divide 19 Jun 2011, 18:53
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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 ? :)
Show more
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.
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In how many different ways can a group of 9 people be divide 19 Jun 2011, 19:51
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voodoochild
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.
Show more

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

Thanks
Voodoo
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In how many different ways can a group of 9 people be divide 21 Jun 2011, 02:29
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voodoochild
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
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Yes, the groups are distinct so no of combinations is 9!/(4!*3!*2!)
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In how many different ways can a group of 9 people be divide 07 Oct 2012, 10:08
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geturdream
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
Show more
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 ?
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In how many different ways can a group of 9 people be divide 30 May 2013, 15:43
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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
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In how many different ways can a group of 9 people be divide 28 Oct 2013, 12:44
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hemanthp
In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?

A. 280
B. 1,260
C. 1,680
D. 2,520
E. 3,360
Show more

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
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In how many different ways can a group of 9 people be divide 25 Jul 2015, 09:11
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geturdream
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 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 ?
Show more


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.

\(\frac{9C3 * 6C3 * 3C3}{3!} = \frac{84*20*1}{3!} = \frac{1680}{3!} = 280\)
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In how many different ways can a group of 9 people be divide 19 Jul 2017, 01:16
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Bunuel

GENERAL RULE:
1. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\)
2. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is NOT important is \(\frac{(mn)!}{(n!)^m * m!}\)
Show more

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
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In how many different ways can a group of 9 people be divide 19 Jul 2017, 01:34
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Bunuel

GENERAL RULE:
1. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\)
2. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is NOT important is \(\frac{(mn)!}{(n!)^m * m!}\)

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.
Show more


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.
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In how many different ways can a group of 9 people be divide 03 Aug 2017, 05:03
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toughmat
Hi Bunuel
Can u pls explain why u r dividing by 3!
we are not ordering here we are only choosing.

We divide by 3! because of exactly what you said: "we are not ordering here we are only choosing."
When you say, "9C3 * 6C3 * 3C3," what you are doing is that you are choosing 3 people of 9 for group 1, 3 people out of the leftover 6 people for group 2 and the rest of the three people for group 3. You have inadvertently marked the 3 groups as distinct. But if we want to just divide them in 3 groups without any distinction of group 1, 2 or 3, we need to divide this by 3!.
Show more

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! ?
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In how many different ways can a group of 9 people be divide 03 Aug 2017, 05:05
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Bunuel

GENERAL RULE:
1. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\)
2. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is NOT important is \(\frac{(mn)!}{(n!)^m * m!}\)

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.


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.
Show more


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! ?
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In how many different ways can a group of 9 people be divide 22 Sep 2018, 18:59
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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).
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In how many different ways can a group of 9 people be divide 27 May 2019, 05:31
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In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?

A. 280
B. 1,260
C. 1,680
D. 2,520
E. 3,360
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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
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