Division of Objects into Groups: The GMAT sometime tends to surprise test-takers with questions on little-known topics. Although the topic of combinations is a widely known and studied topic, a sub-topic within combinations that is sometimes neglected by a significant number of test takers is that of division of objects into groups.

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?

Solution: The answer is simply 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.

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?

Solution: The answer is NOT 28!/(4!12!12!)

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.

Is that it then? No – the GMAT has one last trick up its sleeve.

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 arrangements of the given 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?

Solution: The answer is now 28! x 3!/(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 3 children, B = group with 5 children, C = group with 20 children

OR

A = group with 5 children, B = group with 3 children, C = group with 20 children

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 groups having 3,5, and 20 children and being given the names A,B, C, and D?

Solution: The answer is now 28! x 4C3 x 3!/(5! x 20!)

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