Permutations and Combinations are used to address **Counting Problems** - the number of possible unique ways of putting things together.

However, GRE and GMAT problems in this area, though harder than the other GMAT problems, still tend to be relatively easy compared to what you would face in say, a Math course.

Key things to remember are:

Let us look at a few examples:

1. What is the number of possible ways we can select a man and a woman from a group of 13 men and 12 women?

A man can be selected in one of 13 ways. Each time a man is selected, a woman can be selected in 12 ways, for a total of 13*12=156 possible pairings.

2. What are the possible number of permutations of the digits of 1225?

1225 has 4 digits of which there are two 2s. So the number of permutations is $(4!)/(2!)$=4*3=12.

These are: 1225, 1252, 1522, 2215, 2125, 2152, 2251, 2512, 2521, 5122, 5212, 5221.

3. How many three digit numbers are such that only one digit is even?

This means you want all numbers of the form abc where a is one of 2,4,6,8, and b and c are not one of 2,4,6,8,0.

For these, a can be selected in 4 ways, b and c can be selected in 5 ways each, for a total of 4*5*5=100 possibilities.

And you also want other numbers of the form abc where only one of b or c is of the form 2,4,6,8,0 and the other two digits are not from that set.

Here, a can be selected in 5 ways from 1,3,5,7,9; b in one of 5 ways from 0,2,4,6,8; and c in one of 5 ways from 1,3,5,7,9. This yields 5*5*5=125 possibles.

Or, a can be selected in 5 ways from 1,3,5,7,9; b in one of 5 ways from 1,3,5,7,9; and c in one of 5 ways from 0,2,4,6,8. This gives 5*5*5=125 possibles.

This gives a total of 350 numbers from the 900 three digit numbers we look at, that meet the specified constraints.

However, GRE and GMAT problems in this area, though harder than the other GMAT problems, still tend to be relatively easy compared to what you would face in say, a Math course.

Key things to remember are:

1. n! represents the factorial of n. n!=1*2*3*...*n. 2. 0! =1 by definition. 3. Combinations refer to the number of ways of assembling things together whereIt follows directly from the above that the number of permutations of n objects is obviously greater than the number of possible combinations.ordering is not important. 4. The number of combinations of N distinct objects taken K at a time is represented as $C(N,K)$ or $NC_k$ and is equal to $(N!)/{K!(N-K)!}$.

5. Permutations refer to the number of ways of arranging things where theordering is important.

6. The number of permutations of N distinct objects taken K at a time is represented by $P(N,K)$ or $NP_k$ and is equal to $(N!)/{(N-K)!}$

7. The number of permutations of N objects taken K at a time when a objects are of one kind, b are of a second kind etc is given by $(P(N,K))/{a!b!...}$

Let us look at a few examples:

1. What is the number of possible ways we can select a man and a woman from a group of 13 men and 12 women?

A man can be selected in one of 13 ways. Each time a man is selected, a woman can be selected in 12 ways, for a total of 13*12=156 possible pairings.

2. What are the possible number of permutations of the digits of 1225?

1225 has 4 digits of which there are two 2s. So the number of permutations is $(4!)/(2!)$=4*3=12.

These are: 1225, 1252, 1522, 2215, 2125, 2152, 2251, 2512, 2521, 5122, 5212, 5221.

3. How many three digit numbers are such that only one digit is even?

This means you want all numbers of the form abc where a is one of 2,4,6,8, and b and c are not one of 2,4,6,8,0.

For these, a can be selected in 4 ways, b and c can be selected in 5 ways each, for a total of 4*5*5=100 possibilities.

And you also want other numbers of the form abc where only one of b or c is of the form 2,4,6,8,0 and the other two digits are not from that set.

Here, a can be selected in 5 ways from 1,3,5,7,9; b in one of 5 ways from 0,2,4,6,8; and c in one of 5 ways from 1,3,5,7,9. This yields 5*5*5=125 possibles.

Or, a can be selected in 5 ways from 1,3,5,7,9; b in one of 5 ways from 1,3,5,7,9; and c in one of 5 ways from 0,2,4,6,8. This gives 5*5*5=125 possibles.

This gives a total of 350 numbers from the 900 three digit numbers we look at, that meet the specified constraints.

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