Permutation combination is considered a hard math topic by many GMAT test takers. It is believed that questions from this topic appear when you score in the higher percentile in the GMAT Quant section. Rest assured that GMAT tests questions from this topic that range from easy to medium level of difficulty. Therefore, a thorough understanding of the basics should set you on the path to cracking questions in the GMAT from this topic.
Wizako's Math Lesson Book in this chapter covers concepts right from absolute basics. Most formulae used are derived after explaining the basis with simple examples and in many cases by listing down the number of possibilities. The chapter includes the following concepts:
Here is a typical solved example in Wizako's GMAT Book from this chapter
Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same when looked at in a mirror. They are called symmetric letters. Other letters in the alphabet are asymmetric letters.
How many three letter computer passwords can be formed (no repetition allowed) with at least one symmetric letter?
There are three possible cases that will satisfy the condition of forming three letter passwords with at least 1 symmteric letter.
Case 1: 1 symmetric and 2 asymmetric
Case 2: 2 symmetric and 1 asymmetric
Case 3: all 3 symmetric
1 symmetric letter can be selected from 11 in 11C1 ways.
2 asymmetric letters can be selected from the remaining 15 letters in 15C2 ways.
Number of ways of selecting 1 symmetric and 2 asymmetric = 11C1 × 15C2
2 symmetric letters can be selected from 11 in 11C2 ways.
1 asymmetric letter can be selected from the remaining 15 letters in 15C1 ways.
Number of ways of selecting 2 symmetric and 1 asymmetric = 11C2 × 15C1
3 symmetric letters can be selected from 11 in 11C3 ways.
The 3 distinct letters chosen can be re arranged in 3! ways
Total number of passwords that can be formed = {(11C1 × 15C2) + (11C2 × 15C1) + 11C3} × 3!
= {11 × \\frac{\text{15 × 14}}{\text{1 × 2}}) + \\frac{\text{11 × 10}}{\text{1 × 2}}) × 15 + \\frac{\text{11 × 10 × 9}}{\text{1 × 2 × 3}}) × 6
= {1155 + 825 + 165} × 6
= 2145 × 6 = 12870
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