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Updated on: 03 Apr 2022, 04:03
Originally posted by GMATBusters on 21 May 2018, 23:07.
Last edited by GMATBusters on 03 Apr 2022, 04:03, edited 2 times in total.
Last edited by GMATBusters on 03 Apr 2022, 04:03, edited 2 times in total.
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CONCEPT- Dividing Objects into Groups - Combinations
Source : https://www.f1gmat.com/gmat-combinations-objects-groups
Division into group is one of the most important as well as confusing sub-topic of Combination questions:
It is often neglected by many test takers. But if one is aiming for 90th or higher percentile in Quantitative section, this section is not to be missed.
Type 1: The number of ways to divide m+n+p objects into three groups
having m,n, and p objects is (m+n+p)!/(m! n! p!)
Example: In how many ways can you divide 28 schoolchildren into three groups having
3,5, and 20 children?
Show SpoilerSolution
28!/(3!5!20!)
This problem type is simple enough. However, the GMAT can try to trick you by asking
you a subtle variant of this type of problem.
Type2: The number of ways to divide m+2n objects into three groups
having m,n, and n objects is (m+2n)!/(m! x n! x n! x (no. of groups having the same number of objects)!)
Example: In how many ways can you divide 28 schoolchildren into three groups having
4, 12, and 12 children?
Show SpoilerSolution:
The answer is NOT 28!/(4!12!12!)
Explanation : Instead, we must divide by 2! to get the answer as 28!/(4!12!12!2!)
Why do we divide by 2! in this case?
The reason is that two groups have the same number of objects to be placed in them.
Therefore whether we select an object for one of these two groups or the other, the
selection is essentially the same. Therefore we must divide by the factorial of the number
of groups of the same size in order to account for the extra counting.
Explanation : Instead, we must divide by 2! to get the answer as 28!/(4!12!12!2!)
Why do we divide by 2! in this case?
The reason is that two groups have the same number of objects to be placed in them.
Therefore whether we select an object for one of these two groups or the other, the
selection is essentially the same. Therefore we must divide by the factorial of the number
of groups of the same size in order to account for the extra counting.
Are we done!! No — the GMAT has one last trick up its sleeve.
Type 3: The number of ways to divide m+n+p objects into three groups having m,n, and p objects, where each group has a specific name
assigned to it, is (m+n+p)! x (number of arrangement possible for the names )!/(m! x n! x p!)
Example: In how many ways can you divide 28 schoolchildren into three groups having 3,5, and 20 children and being given the names A,B, and C?
Show SpoilerSolution:
The answer is now 28! x 3! /(3! x 5! x 20!)
Why do we multiple by 3! Here?
Remember, the three names A,B,C can be assigned to the three groups having 3,5, and 20 children in any way.
For instance, we can have
A = group with 20 children ,
B = group with 5 children,
C =group with 5 children
OR
A = group with 5 children
B = group with 3 children,
C = group with 20 children
OR other possible combinations. In fact the number of possible combinations is 3!
To account for the fact that any of the three names can be assigned to any group, we must multiple with the number of arrangements possible for the names.
Why do we multiple by 3! Here?
Remember, the three names A,B,C can be assigned to the three groups having 3,5, and 20 children in any way.
For instance, we can have
A = group with 20 children ,
B = group with 5 children,
C =group with 5 children
OR
A = group with 5 children
B = group with 3 children,
C = group with 20 children
OR other possible combinations. In fact the number of possible combinations is 3!
To account for the fact that any of the three names can be assigned to any group, we must multiple with the number of arrangements possible for the names.
To round off the discussion, here is a final example:
Example: In how many ways can you divide 28 schoolchildren into three gooups having
3,5, and 20 children and being given the names A,B, C, and D?
Show SpoilerSolution:
28!x4C3x3!/(3!x5!x20!). As the number of arrangement possible for the names (selection from A, B, C, D) of three groups = 4C3 x3!
20 Jul 2018, 07:50
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Type2: The number of ways to divide m+2n objects into three groups
having m,n, and n objects is (m+2n)!/(m! x n! x n! x (no. of groups having the same number of objects)!)
Remark: The result is divided by (no. of groups having the same number of objects)! in order to avoid repetition i.e. false counting.
because the two groups having the same number are identical
So we have to divide the numbers of arrangement made by these identical groups.
See the detailed example of division of dividing 2n items into 2 groups of n items.
Gmatbusters.jpg [ 92.59 KiB | Viewed 9607 times ]
having m,n, and n objects is (m+2n)!/(m! x n! x n! x (no. of groups having the same number of objects)!)
Remark: The result is divided by (no. of groups having the same number of objects)! in order to avoid repetition i.e. false counting.
because the two groups having the same number are identical
So we have to divide the numbers of arrangement made by these identical groups.
See the detailed example of division of dividing 2n items into 2 groups of n items.
Attachment:
Gmatbusters.jpg [ 92.59 KiB | Viewed 9607 times ]
Sandy56
06 Nov 2020, 08:59
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