CONCEPT- Dividing Objects into Groups - Combinations

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?

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?

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.

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?

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.

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?

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!

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