GRE® Permutation Combination Practice

Concept: Sampling without replacement; sampling where order does NOT matter

This GRE quant practice question is a permutation combination problem solving question. A classic example of sampling without replacement and where the order does NOT matter. An nCr question.

Question: In how many ways can 3 students be selected from a group of 12 students to represent a school in the inter school essay competition

  1. 33
  2. 12!
  3. 1320
  4. 220
  5. 36

Video Explanation

Scroll down for explanatory answer text

Explanatory Answer

Use these hints to get the answer

  1. Is it a case of sampling with replacement or sampling without replacement?
  2. Is it an example of sampling where the order matters or one where order does not matter?
  3. Based on your answer to the above two questions arrive at the answer. Susbequently, click on the next tab to check whether your method is right.

What kind of sampling?

Total number of students 12. Students to be selected 3. Let us answer the following two questions about the type of sampling and determine the answer.

Is this an example of sampling with replacement or without replacement?

We need to select 3 students from a class of 12 students.

The first student can be any of the 12 students. The second student to be selected will come from a reduced set of 11 students. Else, there are instances where the student selected first could be selected again - in that case we will not have selected 3 students.

Hence, sampling without replacement

Once we have figured that it is an example of sampling without replacement, select 3 students out of 12 in 12C3 ways.

Is it an example of sampling where the order matters or one in which the order does not matter?

Let us say students A, B, and C have been selected from among the 12.

Is it different from saying students B, C, and A have been selected?

NO. So, if ABC is the same as BCA, order does not matter.
If order does not matter, you should not multiply the answer in the previous step with the number of ways things can be reordered.

So, the answer is 12C3

= \\frac{12!}{(12 - 3)! \times 3!}\\) = \\frac{12!}{9! \times 3!} \\) = \\frac{12 \times 11 \times 10 \times 9!}{9! \times 3 \times 2 \times 1} \\) = \\frac{12 \times 11 \times 10}{3 \times 2 \times 1} \\) = 220 ways.

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